Algebra

The given equations are

x + 2y = 7 … (1)

x – 2y = 1 … (2)

Adding (1) and (2),

⇒ x + 2y + x – 2y = 7 + 1

⇒ 2x = 8

⇒ x = 4

Substituting x = 4 in (1),

⇒ 4 + 2y = 7

⇒ 2y = 7 – 4 = 3

⇒ y =

∴ (4,) is the solution to the given system.

The given equations are

3x + y = 8 … (1)

5x + y = 10 … (2)

Here, the coefficients of y in both equations are numerically equal.

Subtracting (2) from (1),

⇒ (3x + y) – (5x + y) = 8 – 10

⇒ 3x + y – 5x – y = – 2

⇒ – 2x = – 2

⇒ x = 1

Substituting x = 1 in (1),

⇒ 3 (1) + y = 8

⇒ y = 8 – 3

⇒ y = 5

∴ (1, 5) is the solution to the given system.

The given equations are

x + = 4 … (1)

+ 2y = 5 … (2)

(1) becomes 2x + y = 8

(2) becomes x + 6y = 15

Now, (2) × 2 – (1)

⇒ 2x + 12y – (2x + y) = 30 – 8

⇒ 2x + 12y – 2x – y = 22

⇒ 11y = 22

⇒ y = 2

Substituting y = 2 in (1),

⇒ 2x + 2 = 8

⇒ 2x = 6

⇒ x = 3

∴ (3, 2) is the solution to the given system.

The given equations are

11x – 7y = xy … (1)

9x – 4y = 6xy … (2)

Dividing both sides of the equation by xy,

⇒ – = 1 i.e. + = 1 … (3)

⇒ – = 6 i.e. + = 6 … (4)

Let a = and b =.

Equations (3) and (4) become

⇒ – 7a + 11b = 1 … (5)

⇒ – 4a + 9y = 6 … (6)

Now, (6) × 7 – (5) × 4

⇒ – 28a + 63b – ( – 28a + 44b) = 42 – 4

⇒ – 28a + 63b + 28a – 44b = 38

⇒ 19b = 38

⇒ b = 2

Substituting b = 2 in (5),

⇒ – 7a + 11(2) = 1

⇒ – 7a = 1 – 22 = – 21

⇒ a = 3

When a = 3, we have = 3. Thus, x =

When b = 2, we have = 2. Thus, y =

∴ (,) is the solution for the given system.

The given equations are

+ = … (1)

+ = … (2)

Multiplying both sides of the equation with xy,

3y + 5x = 20 i.e. 5x + 3y = 20 … (3)

2y + 5x = 15 i.e. 5x + 2y = 15 … (4)

Subtracting (4) from (3),

⇒ 5x + 3y – (5x + 2y) = 20 – 15

⇒ 5x + 3y – 5x – 2y = 5

⇒ y = 5

Substituting y = 5 in (3),

⇒ 5x + 3 (5) = 20

⇒ 5x = 20 – 15 = 5

⇒ x = 1

∴ (1, 5) is the solution for the given system.

The given equations are

8x – 3y = 5xy … (1)

6x – 5y = – 2xy … (2)

Dividing both sides of the equation by xy,

⇒ – = 5 i.e. + = 5 … (3)

⇒ – = – 2 i.e. + = – 2 … (4)

Let a = and b =.

Equations (3) and (4) become

⇒ – 3a + 8b = 5 … (5)

⇒ – 5a + 6y = – 2 … (6)

Now, (5) × 5 – (6) × 3

⇒ – 15a + 40b – ( – 15a + 18b) = 25 – ( – 6)

⇒ – 15a + 40b + 15a – 18b = 31

⇒ 22b = 31

⇒ b =

Substituting b = in (5),

⇒ – 3a + 8() = 5

⇒ – 3a = 5 – =

⇒ a =

When a =, we have =. Thus, x =

When b =, we have =. Thus, y =

∴ (,) is the solution for the given system.

The given equations are

13x + 11y = 70 … (1)

11x + 13y = 74 … (2)

Adding (1) and (2),

⇒ 24x + 24y = 144

Dividing by 24,

⇒ x + y = 6 … (3)

Subtracting (2) from (1),

⇒ 2x + ( – 2y) = – 4

Dividing by 2,

⇒ x – y = – 4 … (4)

Solving (3) and (4),

⇒ x + y + (x – y) = 6 – 4

⇒ 2x = 2

⇒ x = 1

Substituting x = 1 in (3),

⇒ 1 + y = 6

⇒ y = 5

∴ (1, 5) is the solution to the given system.

The given equations are

65x – 33y = 97 … (1)

33x – 65y = 1 … (2)

Adding (1) and (2),

⇒ 98x – 98y = 98

Dividing by 98,

⇒ x – y = 1 … (3)

Subtracting (1) and (2),

⇒ 32x + 32y = 96

Dividing by 32,

⇒ x + y = 3 … (4)

Solving (3) and (4),

⇒ x – y + (x + y) = 1 + 3

⇒ 2x = 4

⇒ x = 2

Substituting x = 2 in (4),

⇒ 2 + y = 3

⇒ y = 1

∴ (2, 1) is the solution to the given system.

The given equations are

+ = 17 … (1)

+ = … (2)

Let a = and b =.

⇒ 15a + 2b = 17 … (3)

⇒ a + b = … (4)

Now, (3) – (4) × 2

⇒ 15a + 2b – (2a + 2b) = 17 –

⇒ 15a + 2b – 2a – 2b =

⇒ 13a =

⇒ a =

Substituting a = in (4),

⇒ + b =

⇒ b = = 7

When a =, =. Thus, x = 5.

When b = 7, = 7. Thus, y =

∴ (5,) is the solution to the given system.

The given equations are

+ = … (1)

+ = 0 … (2)

Let a = and b =.

⇒ 2a + b = … (3)

⇒ 3a + 2b = 0 … (4)

Now, (3) × 3 – (4) × 2

⇒ 6a + 2b – (6a + 4b) = 1/2 – 0

⇒ 6a + 2b – 6a – 4b = 1/2

⇒ – 2b = 1/2

⇒ b =

Substituting b = in (4),

⇒ 3a + 2() = 0

⇒ 3a = 1/2

⇒ a =

When a =, =. Thus, x = 6.

When b = , =. Thus, y = – 4

∴ (6, – 4) is the solution to the given system.

(i)

The like terms are cancelled.

=3x required solution

(ii)

We know a² – b² = (a-b) (a+b)

So,

Also,

x² + 6x + 8 = x² + 4x + 2x +8

= x(x+4) +2(x+4)

= (x+2)(x+4)

And

x² - 5x – 36 = x² - 9x + 4x – 36

= x(x-9)+4(x-9)

= (x+4)(x-9)

So,

The like terms are cancelled.

Required solution

(iii)

x² - 3x – 10 = x² - 5x + 2x – 10

= x(x-5)+2(x-5)

= (x+2)(x-5)

And,

x² - x - 20 = x² - 5x + 4x - 20

= x(x-5)+4(x-5)

= (x+4)(x-5)

We know the formula a³ + b³ = (a+b)(a² + b² - ab)

So x³ + 8 = x³ + 2³

= (x+2)(x² + 4 - 2x)

The like terms are cancelled.

Required solution

(iv)

We know a² – b² = (a-b) (a+b)

x² – 16 = (x-4) (x+4)

x² – 4 = (x-2) (x+2)

x² - 3x + 2 = x² – 2x – x + 2

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

= (x-1)(x-2)

x² - 2x - 8 = x² – 4x + 2x - 8

= x(x-4)+2(x-4)

= (x-4)(x+2)

We know the formula a³ + b³ = (a+b)(a² + b² - ab)

So x³ + 64 = x³ + 4³

= (x+4)(x² + 16 - 4x)

The like terms are cancelled.

Required solution

(v)

The like terms are cancelled.

Required solution

(vi)

We know a³ - b³ = (a-b) (a² + ab + b²)

The like terms are cancelled.

Required solution.

The like terms are cancelled.

Required solution

The like terms are cancelled.

Required solution.

The like terms are cancelled.

Required solution.

The like terms are cancelled.

Required solution

The like terms are cancelled.

=1 required solution.

The like terms are cancelled.

Required solution,

The like terms are cancelled.

Required solution.

(i)

The like terms are cancelled.

=x² + 2x + 4 required solution.

(ii)

The like terms are cancelled.

(iii)

The like terms are cancelled.

Required solution

(iv)

The like terms are cancelled.

Required solution.

(v)

The like terms are cancelled.

Required Solution

(vi)

Like terms are cancelled

Required solution

(vii)

Like terms are cancelled

(viii)

= 0

Let p(x)= , q(x)=

And the rational expression be r(x),

So according to question,

p(x) = q(x) + r(x)

or, r(x) = p(x) – q(x)

So required rational expression r(x)

Let p(x) = , q(x) =

And rational expression be r(x)

According to question,

p(x) – r(x)=q(x)

or, r(x) = p(x) – q(x)

r(x) =

So required rational expression

P – Q

P² –Q² =(P + Q)(P – Q)

=P – Q

So (∵ P² –Q²=P – Q)

=1

In Square root the power of each term is divided by 2

√(196a⁶b⁸c¹⁰) = √(14² a⁶b⁸c¹⁰)

Square Root = |14a³b⁴c⁵|

289 = 17²

⇒Square Root = √[17²(a–b)⁴ (b–c)⁶]

Square Root = |17(a–b)²(b–c)³|

(x + 11)²–44x

⇒ x² + 22x + 121–44x

⇒ x²–22x + 121 = (x–11)²

√[(x + 11)²–44x

Square root = |x–11|

(x–y)² + 4xy

⇒ x² + y²–2xy + 4xy

⇒ x² + y² + 2xy = (x + y)²

√[(x–y)² + 4xy]

Square Root = |x + y|

Square Root

⇒Square Root

Square Root

Square Root

16x² –24x + 9

The above expression can be rewritten as

(4x)²–2×3×4x + 3²

It is in the form of (a–b)²

= (4x–3)²

Square root = √(4x–3)²

|4x–3|

We factorize each of the above polynomials

x² – 25 = x² – 5²

Since it is in the form of a²–b² = (a–b)(a + b)

⇒ x² – 25 = (x–5)(x + 5) …(i)

x² + 8x + 15 = x² + 5x + 3x + 15

⇒ x² + 8x + 15 = x(x + 5) + 3(x + 5)

⇒ x² + 8x + 15 = (x + 3)(x + 5) …(ii)

x² –2x–15 = x² –5x + 3x–15

⇒ x² –2x–15 = x(x–5) + 3(x–5)

⇒ x² –2x–15 = (x + 3)(x–5) … (iii)

Combining (i), (ii) & (iii) we get

(x² – 25)( x² + 8x + 15)( x² –2x–15) = = (x–5)²(x + 5)²(x + 3)²

Square Root = √[(x–5)²(x + 5)²(x + 3)²]

|(x–5)(x + 5)(x + 3)|

The above expression can be rewritten as

(2x)² + (–3y)² + (–5z)² + 2((–3y)×(2x) + (–5z)×(–3y) + (2x)×(–5z))

The above expression is in the form of

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

So the expression becomes (2x–3y–5z)²

Square Root = √(2x–3y–5z)²

|2x–3y–5z|

The equation can be written as

The above equation is in the form of (a + b)² = a² + b² + 2ab

So it becomes

Square root

We factorize each of the above polynomials

6x² + 5x –6 = 6x² + 9x –4x –6

⇒ 6x² + 5x –6 = 3x(2x + 3)–2(2x + 3)

⇒ 6x² + 5x –6 = (3x–2)(2x + 3) …(i)

6x²–x–2 = 6x²–4x + 3x–2

⇒ 6x²–x–2 = 2x(3x–2) + 1(3x–2)

⇒ 6x²–x–2 = (2x + 1)(3x–2) …(ii)

4x² + 8x + 3 = 4x² + 6x + 2x + 3

⇒ 4x² + 8x + 3 = 2x(2x + 3) + 1(2x + 3)

⇒ 4x² + 8x + 3 = (2x + 1)(2x + 3) …(iii)

Combining (i), (ii) & (iii) we get

(6x² + 5x –6) (6x²–x–2)(4x² + 8x + 3) = (3x–2)²(2x + 3)²(2x + 1)²

Square Root = √ (3x–2)²(2x + 3)²(2x + 1)²

| (3x–2)(2x + 3)(2x + 1)|

We factorize each of the above polynomials

2x² –5x + 2 = 2x²–4x–x + 2

⇒ 2x² –5x + 2 = 2x(x–2)–1(x–2)

⇒ 2x² –5x + 2 = (2x–1)(x–2) …(i)

3x²–5x–2 = 3x²–6x + x–2

⇒ 3x²–5x–2 = 3x(x–2) + 1(x–2)

⇒ 3x²–5x–2 = (3x + 1)(x–2) …(ii)

6x² – x –1 = 6x²– 3x + 2x–1

⇒ 6x² – x –1 = 3x(2x–1) + 1(2x–1)

⇒ 6x² – x –1 = (3x + 1)(2x–1) …(iii)

Combining (i), (ii) & (iii) we get

(2x² –5x + 2) (3x²–5x–2) (6x² – x –1) = (2x–1)²(x–2)²(3x + 1)²

Square Root = √(2x–1)²(x–2)²(3x + 1)²

|(2x–1)(x–2)(3x + 1)|

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

|x²–2x + 3|

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

|2x² + 2x + 1|

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient. To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

|3x²–x + 1|

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

|4x²–3x + 2|

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

Since it is a perfect square the remainder is 0

a = –42, b = 49

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

Since it is a perfect square the remainder is 0

a = 12, b = 9

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

We rearrange the equation in the increasing order of power of x.

The polynomial becomes 36–60x + 109x² + bx³ + ax⁴

Since it is a perfect square the remainder is 0

a = 49, b = –70

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

We rearrange the equation in the increasing order of power of x.

The polynomial becomes 36 + 24x + 40x²–bx³ + ax⁴

Since it is a perfect square the remainder is 0

a = 9, b = –12

(2x + 3)² – 81 = 0

= (2x)² + 2(2x)(3) + 3² – 81 = 0

= 4x² + 12x + 9 –81 = 0

= 4x² + 12x – 72 = 0

Divide by 4 both sides

⇒

= x² + 3x – 18 = 0

= x² + 6x – 3x – 18 = 0

= x (x + 6) – 3 (x + 6) = 0

= (x + 6) (x – 3) = 0

x + 6 = 0 or x – 3 = 0

x = –6 or x = 3

3x² – 5x –12 = 0

= 3x² – 9x + 4x – 12 = 0

= 3x (x – 3) + 4(x – 3) = 0

= (3x + 4) (x – 3) = 0

3x + 4 = 0 or x –3 = 0

3x = –4 or x = 3

or x = 3

√5x² + 2x – 3√5 = 0

= √5x² + 5x – 3x – 3√5 = 0

= √5x (x + √5) – 3(x + √5) = 0

= (√5x – 3) (x + √5) = 0

√5x – 3 = 0 or x + √5 = 0

√5x = 3 or x = –√5

or x = –√5

3(x² – 6) =x (x + 7)–3

= 3x² – 18 = x² + 7x – 3

= 3x² – 18 – x² – 7x + 3 = 0

= 2x² – 7x – 15 = 0

= 2x² – 10x + 3x – 15 = 0

= 2x(x – 5) + 3(x – 5) = 0

= (2x + 3)(x – 5) = 0

2x + 3 = 0 or x – 5 = 0

2x = –3 or x = 5

or x = 5

= 3x² – 8 = 2x

= 3x² – 2x – 8 = 0

= 3x² – 6x + 4x – 8 =0

= 3x (x – 2) + 2(x – 2) = 0

= (3x + 2) (x – 2) = 0

3x + 2 = 0 or x – 2 = 0

3x = –2 or x = 2

or x = 2

=

= 5(x² + 1) = 26x

= 5x² + 5 = 26x

= 5x² –26x + 5 = 0

= 5x² – 25x – x + 5 = 0

= 5x (x – 5) – (x – 5) = 0

= (5x – 1) (x – 5) = 0

5x – 1 = 0 or x – 5 = 0

5x = 1 or x = 5

or x = 5

=

=

= 15(x² + x² + 2x + 1) = 34(x² + x)

= 15(2x² + 2x + 1) = 34(x² + x)

= 30x² + 30x + 15 = 34x² + 34x

= 34x² + 34x – 30x² –30x – 15 =0

= 4x² – 4x – 15 = 0

= 4x² – 10x + 6x – 15 = 0

= 2x(2x – 5) + 3(2x – 5) = 0

= (2x + 3) (2x – 5) = 0

2x + 3 =0 or 2x – 5 = 0

2x = –3 or 2x = 5

or x =

a²b²x² – (a² + b²) x + 1 = 0

= a²b²x² – a²x – b²x + 1 = 0

= a²x (b²x – 1) – (b²x – 1) = 0

= (a²x – 1) (b²x – 1) =0

a²x – 1 = 0 or b²x – 1 = 0

a²x = 1 or b²x = 1

2(x + 1)² – 5(x + 1) = 12

= 2(x² + 2x + 1) – 5x – 5 = 12

= 2x² + 4x + 2 – 5x – 5 –12 = 0

= 2x² – x – 15 = 0

= 2x² – 6x + 5x – 15 = 0

= 2x (x – 3) + 5(x –3) = 0

= (2x + 5) (x – 3) = 0

2x + 5 = 0 or x – 3 =0

2x = –5 or x = 3

x = or x = 3

3(x – 4)² – 5(x – 4) = 12

= 3(x² – 8x + 16) – 5x + 20 = 12

= 3x² – 24x + 48 –5x + 20 – 12 = 0

= 3x² – 29x + 56 = 0

= 3x² – 21x – 8x + 56 = 0

= 3x(x – 7) – 8(x – 7) =0

= (x – 7) (3x – 8) = 0

x – 7 = 0 or 3x – 8 = 0

x = 7 or 3x = 8

x² + 6x – 7 = 0

= x² + 6x = 7

Add 9 on both sides

= x² + 6x + 9 = 7 + 9

= x² + 2(3)(x) + 3² = 16

= (x + 3)² = 16

= x + 3 = √16

= x + 3 = ± 4

x + 3 = 4 or x + 3 = –4

x = 4 – 3 or x = – 4 – 3

x = 1 or x = – 7

x² + 3x + 1 =0

= x² + 3x = –1

Add on both sides

2x² + 5x – 3 = 0

= 2x² + 5x = 3

Add on both sides

4x² + 4bx – (a² – b²) = 0

Divide the whole equation by 4

x² – (√3 + 1)x + √3 = 0

= 5x + 7 = (3x + 2)(x – 1)

= 5x + 7 = 3x(x – 1) + 2 (x – 1)

= 5x + 7 = 3x² – 3x + 2x –2

= 5x + 7 = 3x² – x – 2

= 3x² – x – 2 – 5x – 7 =0

= 3x² – 6x – 9 = 0

Divide whole equation by 3

= x² – 2x – 3 = 0

= x² – 3x + x – 3 = 0

= x (x – 3) + (x – 3) = 0

= (x – 3) (x + 1) = 0

x – 3 = 0 or x + 1 = 0

x = 3 or x = –1

x² – 7x + 12 = 0

(–7)²– 4 (1) (12)

=49 – 48

=1

15x² – 11x + 2 = 0

⇒ a = 15 , b = –11 and c = 2

(–11)²– 4 (15) (2)

=121 – 120

=1

=

⇒ 2(x² + 1) = 5x

⇒ 2x² + 2 = 5x

⇒ 2x² – 5x + 2 = 0

(–5)²– 4 (2) (2)

= 25 – 16

= 9

3a²x² – abx – 2b² = 0

⇒ a = 3a² , b = –ab and c = –2b²

(–ab)²– 4 (3a²) (–2b²)

= a²b² + 24a²b²

=25a²b²

a(x² + 1) = x(a² + 1)

⇒ ax² + a = a²x + x

⇒ ax² + a – a²x – x = 0

⇒ ax² – x(a² + 1) + a = 0

⇒ a = a , b = – a² – 1 and c = a

( –a² – 1 )²– 4 (a)(a)

= a⁴ + 2a² + 1 – 4a²

= (a²)² – 2a² + 1

= (a² – 1 )²

36x² – 12ax + (a² –b²) = 0

⇒ a = 36, b = –12a and c = a² – b²

(–12a)²– 4 (36)(a² – b²)

= 144a² – 144(a² – b²)

= 144a² – 144a² + 144b²

= 144 b²

⇒

⇒

⇒ 3(2x² – 7x + 1) = 10(x² – 3x – 4)

⇒ 6x² – 21x + 3 = 10x² – 30x – 40

⇒ 10x² – 6x² – 30x + 21x – 40 –3 = 0

⇒ 4x² – 9x – 43 = 0

(–9)²– 4 (4) (–43)

= 91 + 688

= 769

a²x² + (a² – b²)x – b² =0

⇒ a = a² , b = a² – b² and c = –b²

(a² – b²)²– 4 (a²)(–b²)

= (a²)² – 2a²b² + (b²)² + 4a²b²

= (a²)² + 2a²b² + (b²)²

= (a² + b²)²

Let x be the required number. Then, the reciprocal is .

⇒ sum of a number and its reciprocal is

⇒

⇒

⇒ 8(x² + 1) = 65x

⇒ 8x² + 8 = 65x

⇒ 8x² – 65x + 8 = 0

⇒ 8x² – x – 64x + 8 = 0

⇒ x(8x – 1) – 8(8x – 1) = 0

⇒ (x – 8) (8x – 1) = 0

x – 8 = 0 or 8x – 1 = 0

x = 8 or 8x = 1

Therefore, the two required numbers are 8 and .

Let ‘x’ be the larger number and ‘y’ be the smaller number.

x² – y² = 45 …(1)

y² = 4x …(2)

Now, put the value of y² in equation (1).

x² – 4x = 45

⇒ x² – 4x – 45 = 0

⇒ x²– 9x + 5x – 45 = 0

⇒ x(x – 9) + 5(x – 9) = 0

⇒ (x + 5)(x – 9) =0

x + 5 = 0 or x – 9 = 0

x = –5 or x = 9

Here positive 9 only admissible. From this we need to find the value of y for that we are going to aplly this value in the second equation.

y² = 4 x

⇒ y² = 4 × 9

⇒ y² = 36

⇒ y = √36

⇒ y = ± 6

Here, positive 6 only admissible.

Therefore, the required numbers are 6 and 9.

Let ‘x’ and ‘y’ are the dimension of the vegetable garden.

Area of rectangle = Length × Width

x × y = 100

we are going to cover the barbed wire for fencing only. So, it must be the perimeter of vegetable garden. Usually perimeter always covers all the four side. Bute here we are going to cover only three sides, because one side of the vegetable garden will act as the compound wall.

x + x + y = 30

⇒ 2x + y = 30

⇒

⇒

⇒ 200 + y² = 30y

⇒ y² – 30y + 200 = 0

⇒ y² – 10y –20y + 200 = 0

⇒ y(y – 10) – 20(y – 10) = 0

⇒ (y – 10)(y – 20) =0

y – 10 = 0 or y – 20 = 0

y = 10 or y = 20

Now we are going to apply these values in to get the values of x.

If y = 10 if y = 20

x = 10 x = 5

Therefore, the required dimensions are 10m and 10m or 20m and 5m.

Length of the rectangular field = 20m

Breadth of the rectangular field = 14m

Let x be the uniform width all around the path.

Length of the rectangular field including the path

= 20 + x + x

= 20 + 2x

Width of the rectangular field including path

= 14 + x + x

= 14 + 2x

Area of path = area of rectangular field including path – area of rectangular field

⇒ 111 = (20 + 2x)(14 + 2x) – (20 × 14)

⇒ 111 = 280 + 40x + 28x + 4x² – 280

⇒ 111 = 68x + 4x²

⇒ 4x² + 68x – 111 = 0

⇒ 4x² + 74x – 6x – 111 = 0

⇒ 2x(2x + 37) – 3(2x + 37) = 0

⇒ (2x + 37)(2x – 3) = 0

2x + 37 = 0 or 2x – 3 = 0

2x = –37 or 2x = 3

x = –18.5 or x = 1.5

Therefore, width of the path = 1.5m

Let x be the usual speed of the train.

Let T1 be the time taken to cover the distance 90 km in the speed x km/hr.

Let T2 be the time taken to cover the distance 90 km in the speed x + 15 km/hr.

By using the given condition

Taking 90commonly from two fractions

⇒

⇒

⇒

⇒ 15 × 180 = x² + 15x

⇒ x² + 15x = 2700

⇒ x² + 15x – 2700 = 0

⇒ x² + 60x –45x – 2700 = 0

⇒ x(x + 60) –45(x + 60) = 0

⇒ (x – 45)(x + 60) = 0

x – 45 = 0 or x + 60 = 0

x = 45 or x= –60

therefore speed of the train is 45 km/hr.

Let x km/hr be the speed of water

Speed of boat is 15km/hr.

So, speed in upstream = (15 + x) km/hr.

speed in downstream = (15 – x) km/hr.

Let T1 be the time taken to cover the distance 30 km in upstream.

Let T2 be the time taken to cover the distance 30 km in downstream.

T1 + T2 = 4hours 30minutes

⇒

⇒

⇒

⇒

⇒

⇒

⇒ 900 × 2 = 9(225 – x²)

Now, let us divide the entire equation by 9.

So, that we will get,

200 = 225 – x²

200 + x² = 225

x² = 225 – 200

x² = 25

x = √25

x = ± 5

Speed must be positive so x =5 is the requied speed.

Speed of water = 5km/hr.

Let x be the present age of son

Let ‘y’ be the present age of father

So, x – 1 be the age of son one year age

y – 1 be the age of father one year ago.

By using the given information

y = x²

y – 1 = 8(x – 1)

⇒ y = 8x – 8 + 1

⇒ y = 8x – 7

⇒ x² = 8x – 7

⇒ x² – 8x + 7 = 0

⇒ x² –x – 7x + 7 = 0

⇒ x(x – 1) –7(x – 1) = 0

⇒ (x – 1)(x – 7) = 0

X – 1 = 0 or x – 7 = 0

x = 1 or x = 7

Therefore, age of father is 49.

Let x be the side length of the square board

Area of one square in the chess board = 6.25cm²

Area of 64square = 64×6.25

(x – 4)² = 400

⇒ x – 4 = √400

⇒ x – 4 = ± 20

x – 4 = 20 or x – 4 = –20

x = 20 + 4 or x = –20 + 4

x = 24 or x = –16

Therefore, side length of square shaped chess board is 24cm.

Let x be the tune taken by A to finish the work.

So, x – 6 be the time taken by B to finish the work.

Work done by A in one day =

Work done by B in one day =

Number of days taken by both to finish the work =

⇒

⇒

⇒ 4(2x – 6) = x² – 6x

⇒ 8x – 24 = x² – 6x

⇒ x² – 6x – 8x + 24 = 0

⇒ x² – 14x + 24 = 0

⇒ x² – 12x – 2x + 24 = 0

⇒ x(x – 12) –2(x – 12) = 0

⇒ (x – 12)(x – 2) = 0

x – 12 = 0 or x – 2 = 0

x = 12 or x = 2

Here, 2 is not admissible.

So, B is taking 12 days to finish the work.

Let x km/hr. be the speed of second train.

So, speed of first train will be (x + 15) km/hr.

Distance covered by first train in 2 hours = 2(x + 5)

Distance covered by the second train in 2 hours = 2x

By using Pythagoras theorem

[2(x + 5)]² + (2x)² = 50²

⇒ (2x + 10)² + (2x)² = 50²

⇒ (4x² + 100 + 40x) + 4x² = 2500

⇒ 8x² + 40x + 100 = 2500

⇒ 8x² + 40x + 100 –2500 = 0

⇒ 8x² + 40x – 2400 = 0

Divide by 8 both sides

⇒ x² + 5x – 300 = 0

⇒ x² + 20x – 15x – 300 = 0

⇒ x(x + 20) – 15 (x + 20) = 0

⇒ (x – 15)(x + 20) = 0

x – 15 = 0 or x + 20 = 0

x = 15 or x = –20

Therefore, speed of the second train is 15 km/hr.

x² – 8x + 12 = 0

⇒ a = 1 , b = –8 and c = 12

= (–8)²– 4(1)(12)

64 – 48

= 16

∴ b² – 4ac > 0. hence, roots are real.

2x² – 3x + 4 = 0

⇒ a = 2, b = –3 and c = 4

= (–3)²– 4(2)(4)

9 – 32

= –23

∴ b² – 4ac < 0. hence, roots are not real.

9x² – 12x + 4 = 0

⇒ a = 9, b = –12 and c = 4

= (–12)²– 4(9)(4)

144 – 144

= 0

∴ b² – 4ac > 0. hence, roots are real and equal.

3x² – 2√6 x + 2 =

⇒ a = 3, b = –2√6 and c = 2

= (–2√6 )²– 4(3)(2)

24 – 24

= 0

∴ b² – 4ac = 0. hence, roots are real and equal.

⇒

⇒ 9x² – 60x + 15 = 0

⇒ a = 9, b = –60 and c = 15

= (–60)²– 4(9)(15)

3600 – 540

= 3060

∴ b² – 4ac > 0. hence, roots are real.

(x – 2a)(x – 2b) = 4ab

⇒ x(x –2b) – 2a(x – 2b) = 4ab

⇒ x² – 2bx – 2ax + 4ab – 4ab = 0

⇒ x² – 2x(b + a) = 0

⇒ a = 1, b = –b – a and c = 0

= (–b – a )²– 4(1)(0)

b² + a² + 2ab

∴ b² – 4ac > 0. hence, roots are real.

2x² – 10x + k = 0

∴

⇒ 100 – 8k = 0

⇒ 100 = 8k

=

∴

12x² + 4kx + 3 = 0

∴

⇒ 16k² – 144 = 0

⇒ 16k² = 144

⇒

⇒

∴ k = 3 and k = –3

x² + 2k (x – 2) + 5 = 0

⇒ x² + 2kx – 4k + 5 = 0

∴

⇒ 4k² + 16k – 20 =0

Divide by 4

⇒ k² + 4k – 5 = 0

⇒ k² + 5k – k – 5 = 0

⇒ k(k + 5) – (k + 5) =0

⇒ (k + 5)(k – 1) = 0

k + 5 = 0 or k – 1 = 0

k = –5 or k = 1

∴ k = –5 and k = 1

(k + 1) x² – 2 (k – 1) x + 1 = 0

∴

⇒ 4k² – 8k + 4 – 4k – 4 = 0

⇒ 4k² – 12k = 0

Divide by 4

⇒ k² – 3k = 0

⇒ k(k – 3) = 0

⇒ k = 0 or k– 3 = 0

k = 0 or k = 3

∴ k = 0 and k = 3

x² + 2(a + b) x + 2 (a² + b²) = 0

Compare this equation with ax² + bx + c = 0

∴ a = 1, b = 2(a + b) and c = 2(a² + b²)

b² – 4ac = (2a + 2b)² – 4(1)[2(a² + b²)]

= 4a² + 8ab + 4b² – 8a² – 8b²

= 8ab – 4a² – 4b²

= – 4a² + 8ab – 4b²

= –4(a² – 2ab + b²)

= –4 (a – b)²

Since squared quantity is always positive.

Hence, (a – b)² ≥ 0

Now, it is given a ≠ b, so (a – b)² > 0

So, D = –4(a – b)² will be negative.

Hence the equation has no real roots.

3p²x² – 2pqx + q² = 0

Compare this equation with ax² + bx + c = 0

∴ a = 3p² , b = 2pq and c = q²

b² – 4ac = (2pq)² – 4(3p²)[q²]

= 4p²q² – 12p²q²

= – 8p²q²

Since squared quantity is always positive.

Hence, p²q² ≥ 0

Now, it is given p ≠ q, so p²q² > 0

So, D = –8p²q² will be negative.

Hence the equation has no real roots.

Given: (a² + b²) x² – 2 (ac + bd) x + c² + d² = 0

To prove:

Proof:

We know that,

D = b² – 4ac

If roots are equal, then b² = 4ac

⇒ {–2(ac + bd)}² = 4{(a² + b²)( c² + d²)}

⇒ 4(a²c² + b²d² + 2acbd) = 4 (a²c² + a²d² + b²c² + b²d²)

⇒ 2acbd = a²d² + b²c²

⇒ a²d² + b²c² – 2acbd = 0

⇒ (ad – bc)² = 0

⇒ ad – bc = 0

⇒ ad = bc

⇒

Hence proved.

(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0

⇒ x(x – b) – a(x – b) + x(x – c) – b(x – c) + x(x – a) – c(x – a) = 0

⇒ x² – bx – ax + ab + x² – cx – bx + bc + x² – ax – cx + ac = 0

⇒ 3x² – 2x(a + b + c) + ab + bc + ac = 0

D = b² – 4ac

D = (a + b + c)² – 4(3)(ab + bc + ac) = 0

D = 4(a² + b² + c² + 2ab + 2bc + 2ca – 3ab – 3bc – 3ca)

D = 4(a² + b² + c² – ab – bc – ca)

D = 2[(a –b)² + (b – c)² + (c – a)²]

Which is always greater than zero so the roots are real.

Roots are equal if D = 0

i.e. (a – b)² + (b + c)² + (c – a)² = 0

since sum of three perfect square is equal to zero so each of them separately equal to zero.

So, a – b = 0, b – c = 0, c – a = 0

a = b , b = c, c = a

so, a = b = c.

Given: (1 + m²) x² + 2mcx + c² – a² = 0

To prove: c² = a (1 + m²)

Proof: it is being that equation has equal roots, therefore

D = b² – 4ac = 0 …(1)

From the equation, we have

a = (1 + m²), b = 2mc, c = c² – a²

putting values of a, b and c in (1), we get

D = (2mc)² – 4(1 + m²)(c² – a²)

⇒ 4m²c² – 4(c² + c²m² – a² – a²m²) = 0

⇒ 4m²c² – 4c² – 4c²m² + 4a² + 4a²m² = 0

⇒ –4c² + 4a² + 4a²m² = 0

⇒ 4c² = 4a² + 4a²m²

⇒ c² = a² + a²m²

⇒ c² = a² (1 + m²)

Hence proved.

x² – 6x + 5 = 0

Compare this equation with ax² + bx + c = 0

a = 1, b = –6 and c = 5

kx² + ax + pk = 0

Compare this equation with ax² + bx + c = 0

a = k, b = –a and c = pk

3x² – 5x = 0

Compare this equation with ax² + bx + c = 0

a = 3, b = –5 and c = 0

8x² – 25 = 0

Compare this equation with ax² + bx + c = 0

a = 8, b = 0 and c = –25

i: 3 and 4

Let

∴

∴

∴

∴

ii: 3 + √7 and 3 – √7

Let α=3 + √7 and β=3–√7

∴ α + β=3 + √7 + 3–√7=6 and αβ=(3 + √7)(3–√7)

=9–7=2

∴

∴

∴

iii:

⇒

⇒

3x² – 5x + 2 = 0 compare this with ax² – bx + c = 0

∴ a = 3 , b = –5 and c = 2

i).

⇒

⇒

⇒ =

⇒ =

⇒ =

ii). α – β

α – β = √(α + β)² – 4α β

iii).

3x² – 6x + 4 = 0 compare this with ax² – bx + c = 0

∴ a = 3 , b = –6 and c = 4

2x² – 3x – 5 = 0 compare this with ax² – bx + c = 0

∴ a = 2 , b = –3 and c = –5

Here α = α² and β = β²

General form of quadratic equation whose roots are α² and β²

⇒ x² – (α² + β²) x + α²β² = 0

⇒ x² – (α² + β²) x + (αβ)² = 0

α² + β² = (α + β)² – 2(αβ)

x² – (α² + β²) x + (αβ)² = 0

4x² – 29x + 25 = 0

Therefore the required equation is 4x² – 29x + 25 = 0

x² – 3x + 2 = 0 compare this with ax² – bx + c = 0

∴ a = 1 , b = –3 and c = 2

Here α = –αand β = – β

General form of quadratic equation whose roots are α² and β²

⇒ x² – (–α – β) x + (–α) (–β) = 0

⇒ x² + (α + β) x + (αβ) = 0

⇒ x² + (3)x + (2) = 0

Therefore, the required quadratic equation is x² – 3x + 2 = 0

x² – 3x – 1 = 0 compare this with ax² – bx + c = 0

∴ a = 1 , b = –3 and c = –1

Here

General form of quadratic equation whose roots are α² and β²

⇒

⇒

⇒

= 3²– 2 × (–1)

=9 + 2

= 11

⇒

⇒

⇒ x² – 11x + 1 = 0

Therefore, the required equation is x² + 11x + 1 = 0

3x² – 6x + 1 = 0 compare this with ax² – bx + c = 0

∴ a = 3 , b = –6 and c = 1

i). Here

General form of quadratic equation whose roots are α² and β²

⇒

⇒

⇒

⇒

⇒ x² – 6x + 3 = 0

Therefore, required equation is x² – 6x + 3 = 0

ii). Here, α = α²β and β = β² α

General form of quadratic equation whose roots are α²β and β² α

x² – (α²β + β² α) x + (α²β)(β² α) = 0

⇒ x² – αβ (α + β) x + (α³β³) = 0

⇒ x² – αβ (α + β) x + (αβ)³ = 0

⇒

⇒

⇒

⇒ 27x² – 18x + 1 = 0

Therefore, required equation is 27x² – 18x + 1 = 0

iii). Here, α = 2α + β and β = 2β + α

General form of equation whose roots are 2α + β and 2β + α

x² – (2α + β + 2β + α)x + (2α + β)(2β + α) = 0

x² – (3α + 3β) x + (4αβ + 2α² + 2β² + αβ) = 0

x² – (3α + 3β) x + (2(α² + β² ) + 5αβ) = 0

α² + β² = (α + β)² – 2αβ

x² – (3α + 3β) x + (2(α² + β² ) + 5αβ) = 0

⇒

⇒

⇒

⇒

⇒ 3x² –18x + 70 = 0

Therefore, the required equation is 3x² –18x + 70 = 0

4x² – 3x – 1 = 0 compare this with ax² – bx + c = 0

∴ a = 4 , b = –3 and c = –1

Here

General form of quadratic equation whose roots are α² and β²

⇒

⇒

⇒

⇒

⇒ x² – 3x – 4 = 0

Therefore, required equation is x² – 3x – 4 = 0

Two roots of any quadratic equation are α and β.

Here, one root is square of the other i.e α = β²

3x² – kx – 81 = 0 compare this with ax² – bx + c = 0

∴ a = 3, b = –k and c = –81

αβ = –27

β²(β) = –27

β³ = –27

β³ = (–3)³

β = –3

now, we are going to apply in first equation

⇒

⇒

⇒

⇒ 6 × 3 = k

⇒ k = 18

Roots of any quadratic equation are α and β

Here one root is twice the other.

α = 2β

by comparing the given equation with general form of quadratic equation we get,

a = 2, b = –a and c = 64

_{α + β =}

…(1)

αβ = 32

2β(β) = 32

2β² = 32

⇒

⇒ β² = 16

⇒ β = √16

⇒ β =± 4

⇒

⇒

⇒ 12 × 2 = a

⇒ a = 24

Roots of any quadratic equation are α and β

By comparing the given equation with ax² + bx + c = 0

a = 5, b = –p and c = 1

α – β = 1

α – β = (α + β)² – 4αβ

⇒

⇒

⇒ p² – 20 = 25

⇒ p² = 25 + 20

⇒ p² = 45

⇒ p = √45

⇒ p = 3√5

Given: Two equations: 6x – 2y = 3 and kx – y = 2

Required: To find the value of k such that system of equations have unique solutions

To have unique solution to the System of equations, the required condition is

∴

⇒

⇒

⇒

∴ For every value of k except 3 the system equations have unique solutions.

∴ Correct option is – Option (B)

Given: A system of two linear equations in two variables is consistent

We know that if a system of two linear equations in two variables is consistent then their graph do not intersect at any point

∴ Correct option is – Option (C)

Given: system of equations x –4y = 8 , 3x –12y = 24

Here,

a₁ = 1, b₁ = -4, c₁ = 8 and a₂ = 3, b₂ = -12, c₂ = 24

Now,

, ,

Here, we can clearly see that which is the condition for infinitely many solutions.

∴ The system of equations have infinitely many solutions.

∴ Correct option is – Option (A)

Given: A Quadratic equation p(x) = (k + 4)x² + 13x + 3k

Required: To find the value of k

Let the roots of the given Quadratic equation be: and

∴ Product of roots of the given Quadratic equation is

We know that, Product of roots of a given Quadratic equation is

∴

⇒

⇒ 3k = k + 4

⇒ 2k = 4

⇒ k = 2

∴ The value of k is 2

∴ Correct option is – Option (A)

Given: A Quadratic equation f(x) = 2x² + (p + 3)x + 5 and sum of roots is zero.

Required: to find the value of p

We know that, sum of roots =

∴ = 0

⇒ -(p + 3) = 0

⇒ P + 3 = 0

∴ p = –3

∴ The value of p is –3

∴ Correct option is – Option (C)

Given: x² – 2x + 7

Required: Reminder of x² – 2x + 7 at x + 4

By synthetic division we can find reminder of

x² – 2x + 7 at x = -4

∴ Reminder of x² – 2x + 7 at x + 4 = 31

∴ Correct option is –Option (D)

Given: x² – 2x + 7

Required: Reminder of x³ – 5x² + 7x – 4 at x – 1

By synthetic division we can find Quotient of x³ – 5x² + 7x – 4 at x = 1

∴ Quotient of x² – 2x + 7 when divided by x–1 = x²–4x + 3

∴ Correct option is –Option(B)

Given two polynomials: (x³ + 1) and x⁴ – 1

Required: To find GCD of the given two polynomials

Let f(x) = x³ + 1 and g(x) = X⁴ – 1

Here, degree of g(x) > f(x) ∴ Devisor = x³ + 1

Here reminder = –(x + 1) (Reminder ≠ zero)

∴ The GCD of given two polynomials is -(x + 1)

∴ Correct Option is - Option(C)

Given two polynomials: x² – 2xy + y² and x⁴ – y⁴

Required: To find GCD of the given two polynomials

Let f(x) = x² – 2xy + y² and g(x) = x⁴ – y⁴

f(x) = x² – 2xy + y²

⇒ f(x) = (x–y)²

g(x) = x⁴ – y⁴

⇒ g(x) = (x²)² – (y²)²

⇒ g(x) = (x² – y²)(x² + y²) (∵ a²–b² = (a–b)(a + b))

⇒ g(x) = (x–y)(x + y)(x² + y²) (∵ a²–b² = (a–b)(a + b))

∴ The GCD of given two polynomials is (x–y)

∴ Correct Option is - Option (C)

Given two polynomials: x³ – a³ and (x - a)²

Required: To find LCM of the given two polynomials

Let f(x) = x³ – a³ and g(x) = (x - a)²

here,

f(x) = x³ – a³

⇒f(x) = (x-a)(x² + a² + xa)

g(x) = (x - a)²

⇒g(x) = (x-a)(x-a)

∴ LCM = (x-a) (x² + a² + xa) (x-a) = (x-a)²(x² + a² + xa)

∴ Correct Option is - Option (C)

Given three polynomials: a^k , a^{k + 3} and a^{k + 5}

Required: To find LCM of the given two polynomials

Let f(x) = a^k , g(x) = a^{k + 3} and h(x) = a^{k + 5}

here,

f(x) = a^k

⇒f(x) = a^k

g(x) = a^{k + 3}

⇒g(x) = a^k ×a³

h(x) = a^{k + 5}

⇒h(x) = a^k ×a^{3 + 2} = a^k×a³×a²

∴ LCM = a^k×a³×a² = a^{k + 5}

∴ Correct Option is - Option (D)

Given: Rational Expression:

Required: The lowest form of the given Rational Expression.

Let f(x) =

⇒ f(x) = (∵ factorization)

⇒ f(x) =

⇒ f(x) =

⇒ f(x) =

∴ The lowest form of the given rational expression is:

∴ Correct Option is - Option (B)

Given: Two rational expressions: and

Required: Product of the given two rational numbers

Let f(x) = and g(x) =

Now, f(x)×g(x) =

⇒ f(x)×g(x) =

⇒ f(x)×g(x) =

∴ Product of the given rational expressions is:

∴ Correct Option is - Option (A)

Given: Two polynomials and

Required: Divide by

Let f(x) = and g(x) =

Now,

⇒

⇒

⇒ (∵ a²–b² = (a–b)(a + b))

⇒

∴ Quotient when f(x) is divided g(x) we get is: (x-5)(x-3)

∴ Correct Option is - Option (A)

Given: Two polynomials: and

Required: Two add the given two polynomials

Let f(x) = and g(x) =

Now, f(x) + g(x) = +

⇒ f(x) + g(x) =

⇒ f(x) + g(x) =

⇒ f(x) + g(x) =

⇒ f(x) + g(x) =

⇒ f(x) + g(x) = (∵ a³–b³ = (a–b)(a² + ab + b²))

∴ f(x) + g(x) = a² + ab + b²

That is, the sum of given two polynomials is a² + ab + b²

∴ Correct Option is - Option (A)

Given: A polynomial: 49

Required: to find the square root of the given polynomial

Let f(x) = 49

Now,

⇒ (∵ (x–y)² = x²-2xy + y²)

⇒

⇒

∴ Square root of the given polynomial is:

∴ Correct Option is - Option (D)

Given: A polynomial:

Required: to find the square root of the given polynomial

Let f(x) =

Now,

⇒ (∵ (x–y–z)² = x² + y² + z²-2xy + 2yz-2zx)

⇒

∴ Square root of the given polynomial is:

∴ Correct Option is - Option (D)

Given: A polynomial: 121

Required: to find the square root of the given polynomial

Let f(x) = 121

Now,

⇒

⇒

∴ Square root of the given polynomial is:

∴ Correct Option is - Option (D)

Given: The quadratic equation has equal roots

Here,

b²–4ac = 0 (∵ The quadratic equation has equal roots)

⇒ b²–4ac = 0

⇒ b² = 4ac

⇒ 4ac = b²

⇒ c =

∴ The value of c =

∴ Correct Option is - Option (B)

Given: The quadratic equation and has no real roots.

Required: To find the value of k

Here,

b²–4ac<0 (∵ Quadratic equation has no real roots)

⇒ (5k)²–4(1)(16)<0

⇒ 25k²<64

⇒ k²<

Applying Square root on both sides

⇒ √k² <

⇒

∴ The value of k is

∴ Correct Option is - Option (C)

Given: One of the root of the Quadratic Equation that is 3

Required: To find the Quadratic equation, which satisfies the given root.

Now,

We substitute the zero in every equation given in the options

∴ Case (i): x²–6x-5 = 0

⇒ (3)²–6(3)–5 = 0

⇒ 9–18–5 = –14≠0

Case (ii): x² + 6x–5 = 0

⇒ (3)² + 6(3)–5 = 0

⇒ 9 + 18–5 = 22≠0

Case(iii): x²–5x–6 = 0

⇒ (3)²–5(3)–6 = 0

⇒ 9–15–6 = –12≠0

Case(iv): x²–5x + 6 = 0

⇒ (3)²–5(3) + 6 = 0

⇒ 9–15 + 6 = 0

∴ The Quadratic equation with one root as 3 is x²–5x–6 = 0

∴ Correct Option is - Option (D)

Given: Two Quadratic Equations and

Required: to find the common root of the given Quadratic Equations

Let the common root be α

Α is the root of x²–bx + c = 0

∴ x²–bx + c = 0

⇒ α²–bα + c = 0 -eq(1)

Also, α is the root of x² + bx–a = 0

∴ α² + bα–a = 0 -eq(2)

Here, eq(1) = eq(2)

∴ α² + bα–a = α²–bα + c

⇒ 2bα = c + a

⇒ α =

∴ The common root is α =

∴ Correct Option is - Option(A)

Given: are the roots of

Required: To find the wrong statement in the given options

We know that sum of roots is given by :

∴

We, also know that product of roots is given by:

∴

Now,

⇒

⇒

⇒

⇒

Now,

∴ We can clearly see in the option that is wrong.

∴ Correct Option is - Option (C)

Given: and are roots of

Required:- Quadratic equation with roots and

Sum of roots of given quadratic equation =

∴ = -eq(1)

Product of roots of given quadratic equation =

∴ = -eq(2)

Sum of roots of required quadratic equation =

Product of roots of required quadratic equation =

Here,

Dividing eq(1) by eq(2) we get,

∴ Sum of roots of the required quadratic equation =

Again by making the reciprocal of eq(2), we get

∴ Product of roots of the required quadratic equation =

We know that, when roots of the quadratic equation are known, we can calculate the quadratic equation as:

x²-(sum of roots)x + (product of roots) = 0

∴ Required quadratic equation: x² –() + () = 0

⇒ = 0

⇒ cx² + bx + a = 0

∴ Required quadratic equation is: cx² + bx + a = 0

∴ Correct option is -Option(C)

Given: b = a + c and the quadratic equation has equal roots

Here,

b²–4ac = 0 (∵ The quadratic equation has equal roots)

⇒ b²–4ac = 0

⇒ (a + c)²–4ac = 0 (∵ b = a + c)

⇒ a² + 2ac + c²–4ac = 0

⇒ a²–2ac + c² = 0

⇒ (a–c)² = 0

Applying sq.rt on both sides

⇒ √(a–c)² = √0

⇒ a–c = 0

⇒ a = c

∴ a = c

∴ Correct option is -Option (A)

The given system of equations is

3x + 4y – 24 = 0 and

20x – 11y – 47 = 0

For cross multiplication method, we write the coefficients as

Hence, we get = =

⇒ = =

⇒ = =

⇒ x = = 4

⇒ y = = 3

∴ (4, 3) is the solution to the given system.

The given equations are

0.5x + 0.8y – 0.44 = 0

0.8x +0.6y – 0.5 = 0

For cross multiplication method, we write the coefficients as

Hence, we get = =

⇒ = =

⇒ = =

⇒ x = = 0.4

⇒ y = = 0.3

∴ (0.4, 0.3) is the solution to the given system.

The given equations are

– = – 2 … (1)

+ = … (2)

By taking LCM,

(1) becomes 9x – 10y + 12 = 0

(2) becomes 2x + 3y – 13 = 0

For cross multiplication method, we write the coefficients as

Hence, we get = =

⇒ = =

⇒ = =

⇒ x = = 2

⇒ y = = 3

∴ (2, 3) is the solution to the given system.

The given equations are

– = – 2 … (1)

+ = 13 … (2)

Let a = and b =.

⇒ 5a – 4b + 2 = 0 … (3)

⇒ 2a + 3b – 13 = 0 … (4)

For cross multiplication method, we write the coefficients as

Hence, we get = =

⇒ = =

⇒ = =

⇒ a = = 2

⇒ b = = 3

When a = 2, = 2. Thus, x = 1/2.

When b = 3, = 3. Thus, y =.

∴ (,) is the solution to the given system.

Let x be the greater number and y be the smaller number.

First condition is x = 3y + 2

Equation is x – 3y – 2 = 0 … (1)

Second condition is x = 4y – 5

Equation is x – 4y + 5 = 0 … (2)

For cross multiplication method, we write the coefficients as

Hence, we get = =

⇒ = =

⇒ = =

⇒ x = = 23

⇒ y = = 7

∴ The numbers are 23 and 7.

Let the incomes be x and expenditure be y.

We know that savings = income – expenditure

First condition is

9x – 4y = 2000

Second condition is

7x – 3y = 2000

For cross multiplication method, we write the coefficients as

Hence, we get = =

⇒ = =

⇒ = =

⇒ x = 2000

⇒ y = 4000

∴ Income of first person = 9x = 9 × 2000 = Rs. 18, 000

Income of second person = 7x = 7 × 2000 = Rs. 14, 000

Let x denote the digit in the tenth place and y denote the digit in the unit place. So the number may be written as 10x + y.

When digits are reversed, the number becomes 10y + x.

First condition is

10x + y = 7 (x + y)

⇒ 10x + y – 7x – 7y = 0

⇒ 3x – 6y = 0

⇒ x – 2y = 0

Second condition is

10y + x = (10x + y) – 18

⇒ 10y + x – 10x – y + 18 = 0

⇒ – 9x + 9y + 18 = 0

⇒ – x + y + 2 = 0

For cross multiplication method, we write the coefficients as

Hence, we get = =

⇒ = =

⇒ = =

⇒ x = = 4

⇒ y = = 2

∴ The number is 10x + y = 10 (4) + 2 = 40 + 2 = 42.

Let chairs be x and tables be y.

Then the equations are

3x + 2y = 700 i.e. 3x + 2y – 700 = 0

5x + 3y = 1100 i.e. 5x + 3y – 1100 = 0

For cross multiplication method, we write the coefficients as

Hence, we get = =

⇒ = =

⇒ = =

⇒ x = = 100 (Cost of one chair)

⇒ y = = 200 (Cost of one table)

Now, total cost of two chairs and three tables,

2x + 3y = 2 (100) + 3 (200) = 200 + 600 = Rs. 800

Let length be l and breadth be b.

Then the first condition is

(l + 2) (b – 2) = lb – 28

⇒ lb – 2l + 2b – 4 – lb + 28 = 0

⇒ – 2l + 2b + 24 = 0

⇒ – l + b + 12 = 0

The second condition is

(l – 1) (b + 2) = lb + 33

⇒ lb + 2l – b – 2 – lb – 33 = 0

⇒ 2l – b – 35 = 0

For cross multiplication method, we write the coefficients as

Hence, we get = =

⇒ = =

⇒ = =

⇒ l = = 23

⇒ b = = 11

∴ Area of rectangle= l × b = 23 × 11 = 253 cm²

Let the speed be x km/hr and distance travelled be y km.

We know that distance = speed × time

Scheduled time to cover distance = y/x hr

Then the first condition is

⇒ = – 4

⇒ =

⇒ xy = (x + 6) (y – 4x)

⇒ xy = xy – 4x² + 6y – 24x

⇒ 4x² – 6y + 24x = 0

⇒ 2x² – 3y + 12x = 0 i.e. 12x – 3y + 2x² = 0

Second condition is

⇒ = + 6

⇒ =

⇒ xy = (x – 6) (y + 6x)

⇒ xy = xy + 6x² – 6y – 36x

⇒ 6x² – 6y – 36x = 0

⇒ x² – y – 6x = 0 i.e. – 6x – y + x² = 0

For cross multiplication method, we write the coefficients as

Hence, we get = =

⇒ = =

⇒ = =

⇒ =

⇒ – 30x = – x²

⇒ 30km/hr = x

Now, =

⇒ =

⇒ y = 24 × 30 = 720 km

∴ The distance covered by train = 720 km

Let f(x) = x² – 2x – 8

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ x² – 2x – 8 = 0

To find out roots of this polynomial we use splitting of middle term method.

According to this method we need to find two numbers whose sum is – 2 and product is 8.

∴ x² – (4 – 2)x – 8 = 0

⇒ x² – 4x + 2x – 8 = 0

⇒ x(x – 4) + 2(x – 4) = 0

⇒ (x + 2)(x – 4) = 0

∴ x = – 2 and x = 4.

⇒ Our zeros are α = – 2 and β = 4.

⇒ sum of zeros = α + β = – 2 + 4 = 2.

⇒ Product of zeros = αβ = ( – 2) × 4 = – 8.

⇒ Comparing f(x) = x² – 2x – 8 with standard equation ax² + bx + c = 0.

We get, a = 1, b = – 2 and c = – 8

We can verify,

⇒ Sum of zeros =

i.e. α + β =

∴ α + β = 2

⇒ Product of zeros =

αβ =

αβ = – 8.

Hence, relationship between zeros and coefficient is verified.

Let f(x) = 4x² – 4x + 1

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ 4x² – 4x + 1 = 0

To find out roots of this polynomial we use splitting of middle term method.

According to this method we need to find two numbers whose sum is – 4 and product is 4.

∴ 4x² – (2 + 2)x + 1 = 0

⇒ 4x² – 2x – 2x + 1 = 0

⇒ 2x(2x – 1) – 1(2x – 1) = 0

⇒ (2x – 1)(2x – 1) = 0

∴ 2x – 1 = 0

∴ x = .

Again, 2x – 1 = 0

∴ x =

⇒ Our zeros are α = and β = .

⇒ sum of zeros = α + β = + = 1.

⇒ Product of zeros = αβ = .

Now, Comparing f(x) = 4x² – 4x + 1 with standard equation ax² + bx + c = 0.

We get, a = 4, b = – 4 and c = 1

We can verify,

⇒ Sum of zeros =

i.e. α + β =

∴ α + β = 1

⇒ Product of zeros =

αβ =

Hence, relationship between zeros and coefficient is verified.

Let f(x) = 6x² – 3 – 7x

Arranging equation in proper form.

Now, f(x) = 6x² – 7x – 3

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ 6x² – 7x – 3 = 0

To find out roots of this polynomial we use splitting of middle term method.

According to this method we need to find two numbers whose sum is – 7 and product is – 18.

∴ 6x² – (9 – 2)x – 3 = 0

⇒6x² – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (3x + 1)(2x – 3) = 0

∴ 3x + 1 = 0

∴ x = .

Again, 2x – 3 = 0

∴ x =

⇒ Our zeros are α = and β = .

⇒ sum of zeros = α + β = +

⇒ sum of zeros = α + β =