Polynomials

To find the zeros of the polynomial let us first solve the polynomial by equating it to zero. Factorizing the given polynomial

4x² + 8x = 0

4x(x + 8) = 0

4x = 0 or x + 8 = 0

Now solving the first part,

4x = 0

x = 0/4

x = 0

Now solving the second part,

x + 8 = 0

x = -8

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 4, b = 8, c = 0

Sum of zeroes = -b / a

= -8 / 4

= -2

Product of zeroes = c / a

= 0 / 4

= 0

To find the zeros of the polynomial let us first solve the polynomial by equating it to zero. Factorizing the given polynomial

4x² – 4x + 1 = 0

To factorize the polynomial we have,

Sum of the value should be equal = -4

Product should be equal to = 4 × 1

= 4

So two numbers are -2, -2

4x² – 2x – 2x + 1 = 0

2x(2x – 1) – 1(2x – 1) = 0

(2x-1)(2x-1) = 0

2x-1 = 0 or 2x-1 = 0

Both the parts are same.

Solving them,

2x-1 = 0

2x = 1

x = 1/2

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 4, b = -4, c = 1

Sum of zeroes = -b / a

= - (-4) / 4

= 1

Product of zeroes = c / a

= 1 / 4

= 1/4

To find the zeros of the polynomial let us first solve the polynomial by equating it to zero. Factorizing the given polynomial

6x² – x - 2 = 0

To factorize the polynomial we have,

Sum of the value should be equal = -1

Product should be equal to = 6 × (-2)

= -12

So two numbers are -4, 3

6x² – 4x + 3x - 2 = 0

2x (3x – 2) + 1(3x – 2) = 0

(2x+1)(3x-2) = 0

2x+1 = 0 or 3x-2 = 0

Now Solving first part,

2x+1 = 0

2x = -1

x = -1/2

Now solving the second part,

3x-2 = 0

3x = 2

x = 2/3

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 6, b = -1, c = -2

Sum of zeroes = -b / a

= - (-1) / 6

= 1/6

Product of zeroes = c / a

= -2 / 6

= -1/3

To find the zeros of the polynomial let us first solve the polynomial by equating it to zero. Factorizing the given polynomial

x² – 15 = 0

x² = 15

x = √15

When we take square root on both sides, we get two values of a variable

Thus x = +√15 and x = -√15

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 1, b = 0, c = -15

Sum of zeroes = -b / a

= 0 / 4

= 0

Product of zeroes = c / a

= -15 / 1

= -15

To find the zeros of the polynomial let us first solve the polynomial by equating it to zero. Factorizing the given polynomial

x² – (√3 + 1)x + √3 = 0

To factorize the polynomial we have,

Sum of the value should be equal = -(√3 + 1)

Product should be equal to = 1 × √3

= √3

x² – [2(√3 + 1)x]/2 + √3 = 0

x² - 2(√3 + 1)x/2 + (√3 + 1)²/2² - (√3 + 1)²/2² + √3 = 0

[x - (√3 + 1)/2]² - (3 + 1 + 2√3)/4 + √3 = 0

[x - (√3 + 1)/2]² = (3 + 1 + 2√3 - 4√3)/4

[x - (√3 + 1)/2]² = (√3 - 1)²/2²

[x - (√3 + 1)/2] = ± (√3 - 1)/2

Solving for positive value,

x = (√3 - 1)/2 + (√3 + 1)/2

x = (√3 - 1 + √3 + 1)/2

x = 2√3/2 = √3

Solving for negative value,

x = -(√3 - 1)/2 + (√3 + 1)/2

x = (-√3 + 1 + √3 + 1)/2

x = 2/2 = 1

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 1, b = - (√3 + 1), c = √3

Sum of zeroes = -b / a

= (√3 + 1) / 1

= √3 + 1

Product of zeroes = c / a

= √3 / 1

= √3

To find the zeros of the polynomial let us first solve the polynomial by equating it to zero. Factorizing the given polynomial

3x² – x - 4 = 0

To factorize the polynomial we have,

Sum of the value should be equal = -1

Product should be equal to = 3 × (-4)

= -12

So two numbers are -4, 3

3x² – 4x + 3x - 4 = 0

3x (x + 1) -4 (x + 1) = 0

(3x - 4)(x + 1) = 0

3x - 4 = 0 or x + 1 = 0

Now solving first part,

3x-4 = 0

3x = 4

x = 4/3

Now solving the second part,

x + 1 = 0

x = -1

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 3, b = -1, c = -4

Sum of zeroes = -b / a

= - (-1) / 3

= 1/3

Product of zeroes = c / a

= -4 / 3

= -4/3

Let the two zeroes be a and b.

The generalized form of the quadratic equation with sum and product of zeroes a and b is as follows:

ax² + bx + c = 0 ………………… (i)

Now in this case

∴ a + b = -b/a = -3

∴ ab = c / a = 2

If a = k where k is any real number

b = 3k ………………… (ii)

c = 2k ………………… (iii)

Put values from (ii) and (iii) in (i)

kx² + 3kx + 2k = 0

k (x² + 3x + 2) = 0

Therefore the quadratic equation is as follows:

x² + 3x + 2 = 0

Let the two zeroes be a and b.

The generalized form of the quadratic equation with sum and product of zeroes a and b is as follows:

ax² + bx + c = 0 ………………… (i)

Now in this case

∴ a + b = -b/a = √2

∴ ab = c/a = 1/3

If a = k, where k is any real number

b = -√2k ………………… (ii)

c = k/3 ………………… (iii)

Put values from (ii) and (iii) in (i)

kx² - √2kx + k/3 = 0

3kx² - 3√2kx + k = 0

k (x² - 3√2x + 1) = 0

Therefore the quadratic equation is as follows:

(x² - 3√2x + 1) = 0

Let the two zeroes be a and b.

The generalized form of the quadratic equation with sum and product of zeroes a and b is as follows:

ax² + bx + c = 0 ………………… (i)

Now in this case

∴ a + b = -b/a = -1/4

∴ ab = 1/4 ………………… (iii)

If a = k, where k is any real number

b = k/4………………… (ii)

c = k/4 ………………… (iii)

Put values from (ii) and (iii) in (i)

kx² + k/4x + k/4 = 0

4kx² + kx + k = 0

k (4x² + x + 1) = 0

Therefore the quadratic equation is as follows:

(4x² + x + 1) = 0

Let the two zeroes be a and b.

The generalized form of the quadratic equation with sum and product of zeroes a and b is as follows:

ax² + bx + c = 0 ………………… (i)

Now in this case

∴ a + b = -b/a = 0

∴ ab = c/a = √5

If a = k, where k is any real number,

b = 0 ………………… (ii)

c = √5k ………………… (iii)

Put values from (ii) and (iii) in (i)

kx² + (0)x + (√5k) = 0

k (x² + √5) = 0

Therefore the quadratic equation is as follows:

x² + √5 = 0

Let the two zeroes be a and b.

The generalized form of the quadratic equation with sum and product of zeroes a and b is as follows:

ax² + bx + c = 0 ………………… (i)

Now in this case

∴ a + b = -b/a = 4 ………………… (ii)

∴ ab = c/a = 1 ………………… (ii)

If a = k, where k is any real number

b = -4k

c = k

Put values from (ii) and (iii) in (i)

kx² – 4kx + k = 0

k (x² – 4x + 1) = 0

Therefore the quadratic equation is as follows:

(x² – 4x + 1) = 0

Let the two zeroes be a and b.

The generalized form of the quadratic equation with sum and product of zeroes a and b is as follows:

ax² + bx + c = 0 ………………… (i)

Now in this case

∴ a + b = -b/a = 1

∴ ab = c/a =1

If a = k, where k is any real integer,

b = -k ………………… (ii)

c = k ………………… (iii)

Put values from (ii) and (iii) in (i)

kx² – kx + k = 0

k (x² – x + 1) = 0

Therefore the quadratic equation is as follows:

(x² – x + 1) = 0

Let the two zeroes be a and b.

Given:

Sum of squares of zeroes is 40

a² + b² = 40

The generalized form of the quadratic equation with sum and product of zeroes a and b is as follows:

x² – (a + b)x + ab = 0 ………………… (i)

a + b = 8

ab = k

We also know that,

(a + b)² = a² + b² + 2ab

a² + b² = (a + b) ² – 2ab

(a + b) ² – 2ab = 40

(8)2 – 2k= 40

2k = 64 – 40

2k = 24

k = 12

Therefore the value of k = 12

The solution is as follows:

Therefore the quotient and remainder are as follows:

Quotient = 3x – 5

Remainder = 9x + 10

The solution is as follows:

Quotient = x – 3

Remainder = 7x – 9

The solution is as follows:

Quotient = x² – 8x + 27

Remainder = -60

The solution is as follows:

Quotient = 3x² – x

Remainder = -x + 4

The solution is as follows:

Since the remainder is zero, it is proved that the first polynomial is the factor of second.

The solution is as follows:

Since the remainder is zero, it is proved that the first polynomial is the factor of second.

The solution is as follows:

Since the remainder is not zero, it is proved that the first polynomial is not the factor of second.

Remainder is 9x – 3

Given that √2 and -√2 are the zeroes of the given polynomial.

So (x - √2) (x + √2) = x² - 2

The above relation is same as (a + b) (a – b) = a² – b²

Here a = x and b = √2

The other zeroes are as follows:

2x² – 3x + 1 = 0

Solving the above quadratic equation.

Sum = -3

Product = 2

So the numbers which satisfy the above condition are -2 and -1

2x² – 2x - x + 1 = 0

2x(x – 1) – 1(x – 1) = 0

(2x – 1) (x – 1) = 0

Solving the first part,

2x-1 = 0

2x = 1

x = 1/2

Solving the second part,

x – 1 = 0

x = 1

Therefore the other zeroes of the polynomials are 1/2 and 1.

Given that 2 + √3 and 2 - √3 are the zeroes of the given polynomial.

So (x – (2 + √3)) (x – (2 - √3)) = (x – 2 - √3) (x – 2 + √3)

= x² – 2x + √3x – 2x + 4 – 2√3 - √3x + 2√3 – 3

= x² – 4x + 1

The other zeroes are as follows:

x² – 2x - 35 = 0

Solving the above quadratic equation.

Sum = -2

Product = -35

So the numbers which satisfy the above condition are -7 and 5

x² – 7x + 5x - 35 = 0

x(x – 7) + 5(x – 7) = 0

(x + 5) (x – 7) = 0

Solving the first part,

x + 5 = 0

x = -5

Solving the second part,

x – 7 = 0

x = 7

Therefore the other zeroes of the polynomials are -5 and 7.

Given that -2 is the zero of the polynomial.

So x = -2 which is x + 2 = 0

The other zeroes are as follows:

x² + 11x + 10 = 0

Solving the above quadratic equation.

Sum = 11

Product = 10

So the numbers which satisfy the above condition are 10 and 1

x² + 10x + x + 10 = 0

x(x + 1)+ 10(x + 1) = 0

(x + 10) (x + 1) = 0

Solving the first part,

x + 10 = 0

x = -10

Solving the second part,

x + 1 = 0

x = -1

Therefore the other zeroes of the polynomials are -1 and -10.

Given:

Dividend = f(x) = x³ – 3x² + x + 2

Divisor = g(x)

Quotient = x – 2

Remainder = -2x + 4

There is an important relation between dividend, divisor, quotient and remainder which is as follows:

Dividend = Divisor × Quotient + Remainder

x³ – 3x² + x + 2 = g(x) × (x-2) + (4-2x)

g(x) × (x-2) = x³ – 3x² + x + 2 – (4 – 2x)

= x³ – 3x² + x + 2 – 4 + 2x

g(x) × (x-2) = x³ – 3x² + 3x – 2

Therefore g(x) = (x³ – 3x² + 3x – 2) / (x-2)

Therefore the g(x) = x² – x – 1

Let us solve the above equation,

x (x + 1) + 8 = (x + 2) (x – 2)

x² + x + 8 = (x² + 2x – 2x – 4)

x² + x + 8 = x² – 4

x² + x + 8 - x² + 4 = 0

x + 12 = 0

Since on solving the equation we do not get the second power of x which proves that the following set of equation is not quadratic.

Let us solve the above equation,

(x + 2)³ = x³ – 4

x³ + 3x² × 2 + 3x × 4 + 8 = x³ – 4

x³ + 6x² + 12x + 8 = x³ – 4

x³ + 6x² + 12x + 8 - x³ + 4 = 0

6x² + 12x + 12 = 0

6(x² + 2x + 1) = 0

x² + 2x + 1 = 0

Since on solving the equation we do get the second power of x which proves that the following set of equation is quadratic

Let us solve the above equation,

x³ + 3x + 1 = (x – 2)²

x³ + 3x + 1 = x² – 2 × 2 × x + 4 …….. [(a-b) ² = a² – 2ab + b²]

x³ + 3x + 1 = x² – 4x + 4

x³ + 3x + 1 - x² + 4x - 4 = 0

x³ - x² + 7x - 3 = 0

Since on solving the equation we do get the third power of x along with second power which proves that the following set of equation is not quadratic. A quadratic equation contain the highest power of x which is 2.

Let us solve the above equation by equating it to zero,

x² + 1 + x³ = 0 × x

x² + 1 + x³ = 0

Since on solving the equation we do get the third power of x along with second power which proves that the following set of equation is not quadratic. A quadratic equation contain the highest power of x which is 2.

On factorizing the above equation,

2x² – 5x + 3 = 0

2x² – 2x -3x + 3 = 0

2x(x - 1) – 3(x - 1) = 0

(2x - 3) (x - 1) = 0

Solving the first part,

2x - 3 = 0

2x = 3

x = 3/2

Solving the second part,

x – 1 = 0

x = 1

On factorizing the above equation,

9x² – 3x - 2 = 0

9x² + 3x -6x - 2 = 0

3x(3x + 1) – 2(3x + 1) = 0

(3x - 2) (3x + 1) = 0

Solving the first part,

3x - 2 = 0

3x = 2

x = 2/3

Solving the second part,

3x + 1 = 0

3x = -1

x = -1/3

On factorizing the above equation,

√3x² + 10x + 7√3 = 0

√3x² +3x + 7x + 7√3 = 0

√3x(x + √3) + 7(x + √3) = 0

(√3x + 7) (x + √3) = 0

Solving the first part,

√3x + 7= 0

√3x = -7

x = -7/√3

Solving the second part,

x + √3 = 0

x = -√3

On factorizing the above equation,

x² – 8x + 16 = 0

x² -4x -4x + 16 = 0

x(x - 4) – 4(x - 4) = 0

(x - 4) (x - 4) = 0

Here first and second parts are same,

So x – 4 = 0

x = 4

On factorizing the above equation,

(3x – 5) × (x) = 6 × (x² – 3x + 2)

3x² – 5x = 6x² – 18x + 6

6x² – 3x² – 18x + 5x + 12 = 0

3x² - 13x + 12 = 0

3x² - 9x - 4x + 12 = 0

3x(x - 3) -4(x - 3) = 0

(3x – 4)(x - 3) = 0

Solving the first part,

3x – 4 = 0

3x = 4

x = 4/3

Solving the second part,

x – 3 = 0

x = 3

On factorizing the above equation,

100x² – 20x + 1 = 0

100x² - 10x -10x + 1 = 0

10x(10x - 1) – 1(10x - 1) = 0

(10x - 1) (10x - 1) = 0

Here both the parts are same,

So 10x – 1 = 0

10x = 1

x = 1/10

On factorizing the above equation,

3x² – 2√6x + 2 = 0

3x² - √6x -√6x + 2 = 0

√3x(√3x - √2) – √2(√3x - √2) = 0

(√3x - √2) (√3x - √2) = 0

Here both the roots are equal.

So √3x - √2 = 0

√3x = √2

x = √2 / √3

On factorizing the above equation,

x² + 8x + 7 = 0

x² + 7x + x + 7 = 0

x(x + 7) + 1(x + 7) = 0

(x + 7) (x + 1) = 0

Solving the first part,

x + 7= 0

x = -7

Solving the second part,

x + 1 = 0

x = -1

Cross Multiplying in the above equation,

(x + 3) × (2x – 3) = (3x – 7) × (x + 2)

2x² – 3x + 6x – 9 = 3x² + 6x – 7x – 14

2x² + 3x – 9 = 3x² – x – 14

3x² – 2x² – x - 3x – 14 + 9 = 0

x² - 4x – 5 = 0

On factorizing the above equation,

x² + x – 5x – 5 = 0

x(x + 1) – 5(x + 1) = 0

(x – 5) (x + 1) = 0

Solving the first part,

x – 5 = 0

x = 5

Solving the second part,

x + 1 = 0

x = -1

On factorizing the above equation,

abx² + (b² – ac)x – bc = 0

abx² + b²x – acx – bc = 0

bx(ax + b) – c(ax + b) = 0

(bx – c) (ax + b) = 0

Solving the first part,

bx – c = 0

bx = c

x = c/b

Solving the second part,

ax + b = 0

ax = -b

x = -b / a

Now in the above quadratic equation the coefficient of x² is 3. Let us make it unity by dividing the entire quadratic equation by 3.

x² – 5/3x + 2/3 = 0

x² – 5/3x = -2/3

Now by taking half of the coefficient of x and then squaring it and adding on both LHS and RHS sides.

Coefficient of x = 5/3

Half of 5/3 = 5/6

Squaring the half of 5/3 = 25/36

Now the LHS term is a perfect square and can be expressed in the form of (a-b) ² = a² – 2ab + b² where a = x and b = 5/6

On simplifying both RHS and LHS we get an equation of following form,

(x ± A)² = k²

Taking Square root of both sides.

Now taking the positive part,

x = 6/6

x = 1

Now taking the negative part,

x = 4/6

x = 2/3

Now in the above quadratic equation the coefficient of x² is 5. Let us make it unity by dividing the entire quadratic equation by 5.

x² – 6/5x - 2/5 = 0

x² – 6/5x = 2/5

Now by taking half of the coefficient of x and then squaring it and adding on both LHS and RHS sides.

Coefficient of x = 6/5

Half of 6/5 = 6/10

Squaring the half of 6/10 = 36/100

Now the LHS term is a perfect square and can be expressed in the form of (a-b) ² = a² – 2ab + b² where a = x and b = 6/10

On simplifying both RHS and LHS we get an equation of following form,

(x ± A)² = k²

Taking Square root of both sides.

Now taking the positive part,

Now taking the negative part,

Now taking the positive part,

Now in the above quadratic equation the coefficient of x² is 4. Let us make it unity by dividing the entire quadratic equation by 4.

x² – 3/4x + 5/4 = 0

x² – 3/4x = -5/4

Now by taking half of the coefficient of x and then squaring it and adding on both LHS and RHS sides.

Coefficient of x = 3/4

Half of 3/4 = 3/8

Squaring the half of 3/4 = 9/64

Now the LHS term is a perfect square and can be expressed in the form of (a-b) ² = a² – 2ab + b² where a = x and b = 3/8

On simplifying both RHS and LHS we get an equation of following form,

(x ± A)² = k²

It is observed that the term obtained on RHS is a negative term and taking square root of a negative term will give imaginary roots for the given quadratic equation.

Therefore the given quadratic equation does not has real roots.

Now in the above quadratic equation the coefficient of x² is 4. Let us make it unity by dividing the entire quadratic equation by 4.

x² + √3x + 3/4 = 0

x² + √3x = -3/4

Now by taking half of the coefficient of x and then squaring it and adding on both LHS and RHS sides.

Coefficient of x = √3

Half of √3= √3/2

Squaring the half of √3= 3/4

Now the LHS term is a perfect square and can be expressed in the form of (a-b) ² = a² – 2ab + b² where a = x and b = √3/2

On simplifying both RHS and LHS we get an equation of following form,

(x ± A)² = k²

Here RHS term is zero which implies that the roots of the quadratic equation are real and equal.

Taking square root of both sides,

Now in the above quadratic equation the coefficient of x² is 2. Let us make it unity by dividing the entire quadratic equation by 2.

x² + 1/2x - 2 = 0

x² + 1/2 x = 2

Now by taking half of the coefficient of x and then squaring it and adding on both LHS and RHS sides.

Coefficient of x = 1/2

Half of 1/2 = 1/4

Squaring the half of 1/2 = 1/16

Now the LHS term is a perfect square and can be expressed in the form of (a-b) ² = a² – 2ab + b² where a = x and b = 1/4

On simplifying both RHS and LHS we get an equation of following form,

(x ± A)² = k²

Taking Square root of both sides.

Now taking the positive part,

Now taking the negative part,

Now in the above quadratic equation the coefficient of x² is 2. Let us make it unity by dividing the entire quadratic equation by 2.

x² + 1/2 x + 2= 0

x² + 1/2 x = - 2

Now by taking half of the coefficient of x and then squaring it and adding on both LHS and RHS sides.

Coefficient of x = 1/2

Half of 1/2 = 1/4

Squaring the half of 1/2 = 1/16

Now the LHS term is a perfect square and can be expressed in the form of (a-b) ² = a² – 2ab + b² where a = x and b = 1/4

On simplifying both RHS and LHS we get an equation of following form,

(x ± A)² = k²

It is observed that the term obtained on RHS is a negative term and taking square root of a negative term will give imaginary roots for the given quadratic equation.

Therefore the given quadratic equation does not has real roots.

Now in the above quadratic equation the coefficient of x² is 4. Let us make it unity by dividing the entire quadratic equation by 4.

4x² + 4bx – (a² – b²) = 0

x² + bx = (a² – b²)/4

Now by taking half of the coefficient of x and then squaring it and adding on both LHS and RHS sides.

Coefficient of x = b

Half of b = b/2

Squaring the half of b = b/4

Now the LHS term is a perfect square and can be expressed in the form of (a-b) ² = a² – 2ab + b² where a = x and b = b/2

On simplifying both RHS and LHS we get an equation of following form,

(x ± A)² = k²

Taking Square root of both sides.

Now taking the positive part,

x = (a – b) / 2

Now taking the negative part,

x = - (a + b) / 2

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 2

b = -2√2

c = 1

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ (-2√2)² – (4 × 2 × 1)

⟹ (4 × 2) - 8

⟹ 8 – 8

⟹ 0

Since b² – 4ac = 0 the roots are real and equal.

Now let us put the values in the above formula

x = 1/√2 , 1/√2

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 9

b = 7

c = -2

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ (7)² – (4 × 9 × -2)

⟹ 49 – (-72)

⟹ 49 + 72

⟹ 121

Since b² – 4ac = 121 the roots are real and distinct.

Now let us put the values in the above formula

Solving with positive value first,

x = 4 / 18

x = 2/9

Solving with negative value second,

x = -18 / 18

x = -1

Let us solve the above equation by equating it to zero,

x² + 1 = 3x

x² – 3x + 1 = 0

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 1

b = -3

c = 1

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ (-3)² – (4 × 1 × 1)

⟹ 9 – 4

⟹ 5

Since b² – 4ac = 5 the roots are real and distinct.

Now let us put the values in the above formula

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = √2

b = 7

c = 5√2

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ (7)² – (4 × √2 × 5√2)

⟹ 49 – (20 × 2)

⟹ 49 – 40

⟹ 9

Since b² – 4ac = 121 the roots are real and distinct.

Now let us put the values in the above formula

Solving with positive value first,

x = -√2

Solving with negative value second,

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 1

b = 4

c = 5

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ (4)² – (4 × 1 × 5)

⟹ 16 – 20

⟹ -4

Since b² – 4ac = -4 the roots are imaginary.

Therefore the given quadratic equation does not has real roots.

Let us solve the above equation by equating it to zero,

-2 = 3(x² – 2x)

3x² – 6x + 2 = 0

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 3

b = -6

c = 2

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ (-6)² – (4 × 3 × 2)

⟹ 36 – 24

⟹ 12

Since b² – 4ac = 5 the roots are real and distinct.

Now let us put the values in the above formula

Let the two consecutive odd integers be a and a + 2.

Given:

Sum of squares of numbers = 290

(a)² + (a + 2)² = 290

a² + 4a + 4 + a² = 290

2a² + 4a + 4 = 290

2a² + 4a – 290 + 4= 0

2a² + 4a – 286 =0

2(a² + 2a – 143) =0

a² + 2a – 143 =0

On factorizing the above equation,

a² + 2a – 143 =0

Sum = 2

Product = 143

Therefore the two numbers satisfying the above conditions are -11 and 13.

a² + 13a – 11a – 143 =0

a(a + 13) -11(a + 13) = 0

(a – 11) (a + 13) = 0

Solving first part,

a – 11 = 0

a = 11

Solving second part,

a + 13 = 0

a = -13

Given that the numbers are positive. So a = -13 is not possible.

So First number is 11 and consecutive positive odd number is 13.

Let larger number be x and smaller number be y.

Given:

(Large Number)² – (Smaller Number)2 = 45

x² – y² = 45

y² = x² – 45

Also Smaller Number², y² = 4 × Large Number

x² – 45 = 4x

x² – 4x – 45 = 0

On factorizing the above equation,

Sum = -4

Product = -45

Therefore the two numbers satisfying the above conditions are -9 and 5.

x² – 9x + 5x – 45 = 0

x(x – 9) + 5(x – 9) = 0

(x – 9) (x + 5) = 0

Solving first part,

x – 9 = 0

x = 9

Solving second part,

x + 5 = 0

x = -5

If x = 9, then

y² = 4 × 9

y² = 36

y =± 6

If x = -5, then

y² = 4 × -5

y² = -20

Value of y becomes imaginary which is not possible.

So possible values of x and y are 9, 6 or 9, -6.

Let the smaller part be x and the larger part be 16 – x.

Given:

2 × (Larger Part)² = (Smaller Part)² + 164

2 × (16 – x)² = (x)² + 164

2 × (256 – 32x + x²) = x² + 164

512 – 64x + 2 x² = x² + 164

x² – 64x + 512 – 164 = 0

x² – 64x + 348 = 0

On factorizing the above equation,

Sum = -64

Product = 348

Therefore the two numbers satisfying the above conditions are -58 and -6.

x² – 6x - 58x + 348 = 0

x(x – 6) - 58(x – 6) = 0

(x – 6) (x - 58) = 0

Solving first part,

x – 6 = 0

a = 6

Solving second part,

x - 58 = 0

x = 58

Since x is the smaller it cannot be greater than 16. Hence x cannot be 58

So the smaller part is x = 6

So the larger Part = 16 – 6

= 10

Nature of the quadratic equation can be found out using the following formula

b² – 4ac > 0 for real roots

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 2

b = -3

c = 5

⟹ (-3)² – (4 × 2 × 5)

⟹ 9 – 40

⟹ -31

The answer is negative which implies that the quadratic equation does not have real roots that is they have imaginary roots.

Nature of the quadratic equation can be found out using the following formula

b² – 4ac > 0 for real roots

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 2

b = -4

c = 3

⟹ (-4)² – (4 × 2 × 3)

⟹ 16 – 24

⟹ -8

The answer is negative which implies that the quadratic equation does not have real roots that is they have imaginary roots.

Nature of the quadratic equation can be found out using the following formula

b² – 4ac > 0 for real roots

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 2

b = 1

c = -1

⟹ (1)² – (4 × 2 × -1)

⟹ 1 – (-8)

⟹ 1 + 8

⟹ 9

The answer is positive and also a perfect square which implies that the quadratic equation has real and distinct roots.

Nature of the quadratic equation can be found out using the following formula

b² – 4ac > 0 for real roots

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 1

b = -4

c = 4

⟹ (-4)² – (4 × 1 × 4)

⟹ 16 - 16

⟹ 0

The answer is zero which implies that the quadratic equation has real and equal roots.

Nature of the quadratic equation can be found out using the following formula

b² – 4ac > 0 for real roots

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 2

b = 5

c = 5

⟹ (25)² – (4 × 2 × 5)

⟹ 25 - 40

⟹ -15

The answer is negative which implies that the quadratic equation does not have real roots that is they have imaginary roots.

Nature of the quadratic equation can be found out using the following formula

b² – 4ac > 0 for real roots

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 3

b = -2

c = 1/3

⟹ (-2)² – (4 × 3 × 1/3)

⟹ 4 – 4

⟹ 0

The answer is zero which implies that the quadratic equation has real and equal roots.

Let us first solve the above equation,

kx² – 2kx + 6 = 0

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = k

b = -2k

c = 6

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (-2k) ² – (4 × k × 6) = 0

⟹ 4k² – 24k = 0

⟹ 4k(k – 6) = 0

⟹ (k – 6) = 0 or 4k = 0

⟹ k = 6 or k = 0

Let us first solve the above equation,

x² – 2kx + 1x + k² = 0

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 1

b = -2(k+1)

c = k²

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (-2(k+1)) ² – (4 × k² × 1) = 0

⟹ 4(k² + 2k + 1) – 4 k² = 0

⟹ 4k² + 8k + 4 – 4 k² = 0

⟹ 8k + 4 = 0

⟹ 8k = - 4

⟹ k = - 4/8

⟹ k = - 1/2

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 2

b = k

c = 3

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (k) ² – (4 × 2 × 3) = 0

⟹ k² – 24 = 0

⟹ k² = 24

⟹ k = √( 2 × 2 × 6)

⟹ k = ± 2√6

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = k + 1

b = -2(k – 1)

c = 1

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (-2(k – 1)) ² – (4 × (k + 1) × 1) = 0

⟹ 4(k² - 2k + 1) – 4k - 4 = 0

⟹ 4k² - 8k + 4 – 4k - 4 = 0

⟹ 4k² - 12k = 0

⟹ 4k (k – 3) = 0

⟹ (k – 3) = 0 or 4k = 0

⟹ k = 3 or k = 0

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = k + 4

b = k + 1

c = 1

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (k + 1) ² – (4 × (k + 4) × 1) = 0

⟹ k² + 2k + 1 – 4k - 16 = 0

⟹ k² - 2k – 15 = 0

⟹ k² - 2k – 15 = 0

On factorizing the above equation,

Sum = -2

Product = -15

Therefore the two numbers satisfying the above conditions are 3 and -5.

k² – 5k + 3k – 15 = 0

k(k – 5) + 3(k – 5) = 0

(k + 3) (k – 5) = 0

Solving first part,

k + 3 = 0

k = -3

Solving second part,

k – 5 = 0

k = 5

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = k

b = -5

c = k

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (-5) ² – (4 × k × k) = 0

⟹ 25 - 4k² = 0

⟹ 4k² = 25

⟹ k² = 25/4

⟹ k= ± 5/2

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = k

b = 2

c = 1

Since the quadratic equations have real and distinct roots,

b² – 4ac > 0 for real and distinct roots

⟹ (2) ² – (4 × k × 1) > 0

⟹ 4 – 4k > 0

⟹ 4k < 4 [Dividing both sides by 4]

⟹ k < 1

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = k

b = 6

c = 1

Since the quadratic equations have real and distinct roots,

b² – 4ac > 0 for real and distinct roots

⟹ (6)² – (4 × k × 1) > 0

⟹ 36 – 4k > 0

⟹ 4k < 36 [Dividing both sides by 4]

⟹ k < 9

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 1

b = -k

c = 9

Since the quadratic equations have real and distinct roots,

b² – 4ac > 0 for real and distinct roots

⟹ (-k) ² – (4 × 9 × 1) > 0

⟹ k ² – 36 > 0

⟹ k ² > 36

⟹ k > ± 6

Which implies k < -6 and k > 6.

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 1

b = 5k

c = 16

For the quadratic equations to have imaginary [not real] roots,

b² – 4ac < 0 for imaginary [not real] roots

⟹ (5k) ² – (4 × 16 × 1) > 0

⟹ 25k ² – 64 < 0

⟹ 25k ² < 64

⟹ k ² < 64/25

⟹ k <± 8/5

Therefore the range of k values for roots to be imaginary are:

-8/5 < k < 8/5

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = b – c

b = c – a

c = a – b

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (c – a) ² – [4 × (b – c) × (a – b)] = 0

⟹ c² – 2ac + a² – [4 × (ba – b² – ca – bc)] = 0

⟹ c² – 2ac + a² – [4ba – 4b² – 4ca + 4bc] = 0

⟹ c² – 2ac + a² – 4ba + 4b² + 4ca - 4bc = 0

⟹ c² + a² – 4ba + 4b² + 2ac - 4bc = 0

⟹ a² + 4b² + c² – 4ab - 4bc + 2ac = 0

⟹ a² + (-2b)² + c² + 2 × a(-2b) + 2 × (-2b)c + 2 × ac = 0

We have the following formula:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

So according to the formula

⟹ (a + (-2b) + c) ²= 0

Taking square root of both sides

⟹ a + (-2b) + c = 0

∴ 2b = a + c

Hence Proved.

Let u = 24x²yz

and v = 27x⁴y²z²

Writing u and v in factorized form which is as follows:

u = 2 × 2 × 2 × 3 × x² × y × z

= 2³ × 3 × x² × y × z

v = 3 × 3 × 3 × x⁴ × y² × z²

= 3³ × x⁴ × y² × z²

Now selecting the common multiples from both u and v,

= 2³ × 3³ × x⁴ × y² × z²

= 216 x⁴y²z²

Therefore the LCM of u and v = 216 x⁴y²z²

Let u(x) = x² – 3x + 2

On factorizing the above equation,

Sum = -3

Product = 2

Therefore the two numbers satisfying the above conditions are -2 and -1.

u(x) = x² – x – 2x – 2

u(x) = x(x – 1) - 2(x – 1)

u(x) = (x – 2) (x – 1)

Let v(x) = x⁴ + x³ – 6x²

= x²(x² + x – 6)

Solving the quadratic part,

⟹ x² + x – 6

⟹ x² + 3x – 2x – 6

⟹ x(x + 3) – 2(x + 3)

⟹ (x – 2) (x + 3)

Therefore v (x) = x² (x – 2) (x + 3)

By comparing u(x) and v(x),

LCM = x² (x – 2) (x + 3) (x – 1)

Let u(x) = 2x² – 8

On factorizing the above equation,

u(x) = 2(x² – 4)

Let v(x) = x² – 5x + 6

On factorizing the above equation,

Sum = -5

Product = 6

Therefore the two numbers satisfying the above conditions are -2 and -3.

v(x) = x² – 3x – 2x + 6

v(x) = x(x – 3) - 2(x – 3)

v(x) = (x – 2) (x – 3)

By comparing u(x) and v(x),

LCM = 2(x² – 4) (x – 3)

Let u(x) = (x² – 1) = (x + 1) (x – 1)

Let v(x) = (x² + 1) (x + 1)

Let w(x) = x² + x –1

By comparing u(x), v(x) and w(x),

The LCM is = (x + 1) (x – 1) (x² + 1) (x² + x –1)

= (x² – 1) (x² + 1) (x² + x –1)

= (x⁴ – 1) (x² + x –1)

Let u(x) = 18(6x⁴ + x³ – x²)

= 18 x²(6x² + x –1)

Let us solve the inner quadratic equation,

6x² + x –1 = 0

6x² + x - 1 = 0

6x² + 3x - 2x - 1 = 0

3x(2x + 1) - 1(2x + 1) = 0

(3x – 1) (2x + 1) = 0

So u(x) = 18 x²(3x – 1) (2x + 1)

Let v(x) = 45(2x⁶ + 3x⁵ + x⁴)

= 45 x⁴(2x² + 3x + 1)

Let us solve the inner quadratic equation,

2x² + 3x + 1 = 0

2x² + 2x + x + 1 = 0

2x(x + 1) + 1(x + 1) = 0

(2x + 1) (x + 1) = 0

So v(x) = 45 x⁴(2x + 1) (x + 1)

By comparing the above equation, we get

LCM = 90 x⁴(2x + 1) (x + 1) (3x – 1)

(i) Let u(x) = a³b⁴

Let v(x) = ab⁵

Let w(x) = a²b⁸

By comparing all the above equations, we get,

HCF = ab⁴ (Least power of a and b)

(ii) Let u(x) = 2⁴ × x² × y²

Let v(x) = 2⁴ × 3 × x⁴ × z

By comparing all the above equations, we get,

HCF = 2⁴ × x²

= 16 x² (Least power of x, y and z and also common terms of u(x) and v(x))

(iii) Let u(x) = x² – 7x + 12

= x² – 4x – 3x + 12

= x(x – 4) – 3(x – 4)

= (x – 3) (x – 4)

Let v(x) = x² – 10x + 21

= x² – 7x – 3x + 21

= x(x – 7) – 3(x – 7)

= (x – 7) (x – 3)

Let w(x) = x² + 2x – 15

= x² + 5x – 3x – 15

= x(x + 5) – 3(x + 5)

= (x – 3) (x + 5)

By comparing all the above equations, we get,

HCF = (x + 3) [only common term from u(x), v(x) and w(x)].

(iv) Let u(x) = (x + 3)² (x – 2)

Let v(x) = (x + 3) (x – 2)²

By comparing all the above equations, we get,

HCF = (x + 3) (x – 2) [Least power and common term from u(x) and v(x)].

(v) Let u(x) = (8 × 3) (6x⁴ – x³ – 2x²)

= (8 × 3) x² (6x² – x – 2)

= (8 × 3) x² (6x2 – 4x + 3x – 2)

= (8 × 3) x² (2x (3x – 2) + 1(3x – 2))

= (8 × 3) x² (2x + 1) (3x – 2)

Let v(x) = 20(6x⁶ + 3x⁵ + x⁴)

= (4 × 5) x⁴(6x² + 3x + 1)

= (4 × 5) x⁴(6x² + 3x + 1)

By comparing all the above equations, we get,

HCF = 4 x² (2x + 1) [Least power and common term from u(x) and v(x)].

u(x) = (x – 1)²

= (x – 1) (x – 1)

v(x) = (x² – 1)

= (x + 1) (x – 1)

LCM of u(x) and v(x) = (x – 1)² (x + 1)

HCF of u(x) and v(x) = (x – 1)

u(x) × v(x) = (x – 1) (x – 1) × (x² – 1)

= (x² – 2x + 1) × (x² – 1)

= x⁴ – 2x³ + x² - x² + 2x – 1

= x⁴ – 2x³ + 2x – 1

HCF × LCM = (x – 1)² (x + 1) × (x – 1)

= (x² – 2x + 1) (x² – 1)

= x⁴ – 2x³ + x² - x² + 2x – 1

= x⁴ – 2x³ + 2x – 1

So it is observed that HCF × LCM = u(x) × v(x).

Hence Proved.

Product = (x – 7) (x² + 8x + 12)

= x³ + 8x² + 12x – 7x² – 56x – 84

= x³ + x² – 44x - 84

The LCM = Product / HCF

= (x³ + x² – 44x - 84) / (x + 6)

The division is as follows:

HCF = (x – 5)

LCM = x³ – 19x – 30

= x³ – 19x – 38 + 8

= x³ + 8 – 19x – 38

= x³ + 2³ – 19(x + 2)

x³ + 2³ = (x + 2) (x² – 2x + 4) [Using a³ + b³ = (a + b) (a² – ab + b²)]

= (x + 2) (x² – 2x + 4) – 19(x + 2)

= (x + 2) (x² – 2x + 4 – 19)

= (x + 2) (x² – 2x – 15)

= (x + 2) (x² – 5x + 3x – 15)

= (x + 2) (x(x– 5) + 3(x – 5))

= (x + 2) (x – 5) (x + 3)

Since HCF = x – 5 it will belong to both polynomials.

So u(x) = (x – 5) (x + 3)

= x² – 2x - 15

So v(x) = (x – 5) (x + 2)

= x² – 3x - 10

Let the first root be a.

So the second root as per the question is 1/a.

Product of zeroes = a × 1/a

= 1

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 5, b = 13, c = k

Product of zeroes = c / a

k/5 = 1

Therefore k = 5

So the correct answer is C [5].

Solving the quadratic part,

⟹ x² - x – 6 = 0

⟹ x² - 3x + 2x – 6 = 0

⟹ x(x - 3) + 2(x - 3) = 0

⟹ (x + 2) (x - 3) = 0

Solving first part,

x + 2 = 0

x = - 2

Solving second part,

x - 3 = 0

x = 3

So the correct answer is C [3, -2].

Given one zero = 3

Let second zero a.

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 2, b = 1, c = k

Sum of zeroes = -b/a

3 + a = -1/2

a = -1/2 – 3

a = -7/2

Product of zeroes = 3 × (-7/2)

= -21/2

Product of zeroes = c/a

k/2 = -21/2

k = -21

So the correct answer is D [-21].

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 1

b = -p

c = -p-c

Sum of zeroes = -b/a

= - (-p) / 1

α + β = p …………………………... (i)

Product of Zeroes = c/a

αβ = - (p + c) …………………… (ii)

Now from LHS we have,

(α + 1)(β + 1) = αβ + α + β + 1

From (i) and (ii) we have the values of αβ and (α + β)

(α + 1)(β + 1) = - (p + c) + p + 1

= - p – c + p + 1

= c + 1

Given (α + 1) (β + 1) = 0

c + 1 = 0

c = -1

So the correct answer is B [-1].

x² – kx + 4 = 0

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 1

b = -k

c = 4

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (-k) ² – (4 × 1 × 4) = 0

⟹ k² – 16 = 0

⟹ k² = 16

⟹ k= ± 4

So the correct answer is C [4].

Put x = 1 in the first equation,

a + a + 3 = 0

2a = -3

a = -3/2

Put x = 1 in the second equation,

1 + 1 + b = 0

b + 2 = 0

b = -2

Therefore the value of ab = -3/2 × -2

= 3

So the correct answer is D [3].

The discriminant of the equation can be found out using the following formula:

⟹ b² – 4ac

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 3√3

b = 10

c = √3

⟹ b² – 4ac

⟹ (10)² – (4 × 3√3 × √3)

⟹ (10)² – (4 × 9)

⟹ 100 – 36

= 64

So the correct answer is B [64].

The nature of roots of the equation can be found out using the following formula:

⟹ b² – 4ac

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 4

b = -12

c = -9

⟹ b² – 4ac

⟹ (-12)² – (4 × -9 × 4)

⟹ 144 – (-144)

⟹ 144 + 144

= 288

Since b² – 4ac>0, the roots will be real and distinct.

So the correct answer is B [real and distinct].

Let us write both the expressions in factorized form.

Let u(x) = 2³ × a²× b² × c

Let v(x) = 2² × 5 × a × b³ × c²

So by comparing u(x) and v(x), HCF = 4ab²c [Least power of variables and common terms]

So the correct answer is A [4ab²c].

Let us factorize x² +2x + 1

But x² +2x + 1 is a perfect square and it can be expressed as (x + 1)²

u(x) = x² +2x + 1

u(x) = x² +x + x + 1

u(x) = x(x + 1) +1(x + 1)

u(x) = (x + 1) (x + 1)

Let v(x) = x² – 1

= (x + 1) (x – 1)

So by comparing both polynomials, LCM is = (x1) (x+1)²

So the correct answer is C [(x – 1)(x + 1)²].

Let u(x) = 2 × 3 × x² × y⁴

Let v(x) = 2 × 5 x × x × y²

So by comparing u(x) and v(x), HCF = 2xy² [Least power of variables and common terms]

So the correct answer is B [2xy²].

The generalized formula for the quadratic equation is as follows:

ax² + bx + c = 0

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

The generalized formula for the quadratic equation is as follows:

ax² + bx + c = 0

The discriminant of the equation can be found out using the following formula:

⟹ b² – 4ac

If b² – 4ac > 0, then the roots will be real and distinct.

If b² – 4ac = 0, then the roots will be real and equal.

If b² – 4ac < 0, then the roots will be imaginary.

To find the zeros of the polynomial let us first solve the polynomial by equating it to zero. Factorizing the given polynomial

2x² – 8x + 6 = 0

To factorize the polynomial we have,

Sum of the value should be equal = -8

Product should be equal to = 2 × 6

= 12

So two numbers are -2, -6

2x² – 2x – 6x + 6 = 0

2x(x – 1) – 6(x – 1) = 0

(2x-6)(x – 1) = 0

2x-6 = 0 or x-1 = 0

Solving first part,

2x-6 = 0

2x = 6

x = 3

Solving second part,

x – 1 = 0

x = 1

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 2, b = -8, c = 6

Sum of zeroes = -b / a

= - (-8) / 2

= 4

Product of zeroes = c / a

= 6 / 2

= 3

Zeroes obtained are 3 and 1 and their sum is 4 and product is 3 which is matching to the answer obtained through the ratio of coefficients.

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 1

b = -p

c = q

Sum of zeroes = -b / a

= - (-p) / 1

α + β = p

Product of zeroes = c / a

= q / 1

αβ = q

(i) α² + β²

(α + β) ² = α² + β² + 2αβ

α² + β² = (α + β) ² - 2αβ

= (p) ² – 2q

= p² – 2q

(ii)

The division is as follows:

There is an important relation between dividend, divisor, quotient and remainder which is as follows:

Dividend = Divisor × Quotient + Remainder

Dividend – Remainder = Divisor as the when remainder is subtracted from dividend, the result obtained is completely divisible by divisor.

(-9 + 2k) x + (10 – 8k + k²) = (x + a)

On comparing both sides of coefficients of x:

2k – 9 = 1

2k = 10

k = 5

On comparing both sides:

10 – 8k + k² = a

10 - (8 × 5) + 5² = a

10 – 40 + 25 = a

35 – 40 = a

-5 = a

Hence a = -5 and k = 5.

Let the length be l and breadth be b.

Area = l × b

Given Length = 2b + 1

(2b + 1) × b = 528

2b² + b = 528

2b² + b – 528 = 0

2b² + 33b – 32b – 528 = 0

2b(b – 16) + 33(b – 16) = 0

(2b + 33) (b – 16) = 0

Solving the first part,

2b + 33 = 0

2b = -33

b = -33/2

Solving second part,

b – 16 = 0

b = 16

Breadth cannot be negative.

So breadth = 16 m

Length = 2 × 16 + 1

= 33 m

x² + 4x – 5 = 0

x² + 4x = 5

Now by taking half of the coefficient of x and then squaring it and adding on both LHS and RHS sides.

Coefficient of x = 4

Half of 4 = 2

Squaring the half of 4 = 4

x² + 4x + 4= 5 + 4

(x + 2)² = 9

On simplifying both RHS and LHS we get an equation of following form,

(x ± A)² = k²

Taking square root of both sides,

(x + 2) = ±3

Solving first with positive sign of 3,

x + 2 = 3

x = 3 – 2

x = 1

Solving with negative sign of 3,

x + 2 = -3

x = -3 – 2

x = -5

Let us solve the above equation by equating it to zero,

-2 = 3(x² – 2x)

3x² – 6x + 2 = 0

These problems cannot be solved by normal factorization as it do not contain normal factors.

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ (-6)² – (4 × 3 × 2)

⟹ 36 – 24

⟹ 12

Since b² – 4ac = 12 the roots are real and distinct.

Now let us put the values in the above formula

Cross multiplying,

7× 6 = 6 × (x² – 4x - 5)

42 = 6x² – 24x – 30

6x² – 24x – -30 - 42= 0

6x² – 24x – 72= 0

6(x² – 4x – 12) = 0

x² – 4x – 12 = 0

On factorizing,

x² – 6x + 2x – 12 = 0

x(x – 6) + 2(x – 6) = 0

(x + 2) (x – 6) = 0

Solving first part,

x + 2 = 0

x = -2

Solving second part,

x – 6 = 0

x = 6

Let us solve the above equation by equating it to zero,

x² - 1 = 3x

x² – 3x - 1 = 0

These problems cannot be solved by normal factorization as it do not contain normal factors.

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ (-3)² – (4 × 1 × -1)

⟹ 9 + 4

⟹ 13

Since b² – 4ac = 13 the roots are real and distinct.

Now let us put the values in the above formula

Cross Multiplying,

30 × (-11) = 11 × (x² – 3x - 28)

-330 = 11x² – 33x – 308

11x² – 33x – 308 + 330 = 0

11x² – 33x + 22 = 0

11(x² – 3x + 2) = 0

x² – 3x + 2 = 0

On factorizing,

x² – 2x – x + 2 = 0

x (x – 2) –1(x – 2) = 0

(x – 2) (x – 1) = 0

Solving first part,

x – 2 = 0

x = 2

Solving Second part,

x – 1 = 0

x = 1

Since -5 is the zero of the first equation,

Put x = -5 in first equation

2(-5)² + p (-5) – 15 = 0

2× 25 – 5p – 15 = 0

5p = 50 – 15

5p = 35

p = 7

If the roots are equal, then

b² – 4ac = 0

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = p = 7

b = p = 7

c = k

(7)² – (4 × 7 × k) = 0

(7)² – 28 k = 0

28k = 49

k = 49 / 28

k = 7 / 4

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = p²

b = (p² – q²)

c = – q²

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ ((p² – q²)) ² – (4 × p × – q²)

⟹ (p⁴ –2 p² q² + q⁴) - (-4p²q²)

⟹ p⁴ –2 p² q² + q⁴ + 4p²q²

⟹ (p⁴ + 2 p² q² + q⁴)

⟹ ((p² + q²)) ²

Now let us put the values in the above formula

Solving with positive value first,

x = q² / p²

Solving with negative value second,

x = -1

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 9

b = – 9(a + b)

c = (2a² + 5ab + 2b²)

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ (– 9(a + b)) ² – (4 × 9 × (2a² + 5ab + 2b²))

⟹ 81(a² +2ab + b²) - (36(2a² + 5ab + 2b²))

⟹ 81a² +162ab + 81b² - 72a² - 180ab - 72b²

⟹ 9a² - 18ab + 9b²

⟹ 9(a² - 2ab + b²)

⟹ 3²(a – b) ²

Now let us put the values in the above formula

Solving with positive value first,

Solving with negative value second,

Given:

HCF = x – 1

Let the two polynomials be u(x) and v(x)

So (x – 1) is common to both u(x) and v(x)

LCM = x³ – 7x + 6

= x³ – 7x + 6

= x³ – (1 + 6) x + 6

= x³ – x - 6x + 6

= x(x² – 1) - 6(x - 1)

(x² – 1) = (x + 1) (x – 1)

a² – b² = (a + b) (a – b)

Here a = x and b = 1

= x(x + 1) (x – 1) - 6(x - 1)

= (x-1) [x(x + 1) - 6]

Now solving the inner quadratic equation,

x² + x - 6 = 0

x² + 3x - 2x - 6 = 0

x(x + 3) - 2(x + 3) = 0

(x + 3) (x – 2) = 0

∴ LCM = (x-1) (x + 3) (x – 2)

Since HCF = (x-1), which implies that both u(x) and v(x) contains (x – 1)

Therefore u(x) = (x – 1) (x + 3)

= x² + 2x - 3

v (x) = (x – 1) (x – 2)

= x² – 3x + 2

Therefore the polynomial are x² + 2x – 3 and x² – 3x + 2.

LCM = x³ – 6x² + 3x + 10

= x³ – 5x² - x² + 3x + 10

= x²(x – 5) – (x2 – 3x – 10)

= x²(x – 5) – (x2 – 5x + 2x – 10)

= x²(x – 5) – (x(x – 5) + 2(x – 5))

= x²(x – 5) – (x + 2) (x – 5)

= (x – 5) [x² – x -2]

= (x – 5) [x² – 2x + x -2]

= (x – 5) [x(x – 2) +1(x – 2)]

= (x – 5) (x – 2) (x + 1)

Since HCF = (x + 1) it belongs to both the polynomials.

So u(x) = (x + 1) (x – 5)

= x² – 4x - 5

So v(x) = (x + 1) (x – 2)

= x² – x - 2