Quadratic Equations

Given x² + 1 = 0

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

x² – i² = 0

⇒ (x + i)(x – i) = 0 [∵ a² – b² = (a + b)(a – b)]

⇒ x + i = 0 or x – i = 0

⇒ x = –i or x = i

∴ x = ±i

Thus, the roots of the given equation are ±i.

Given 9x² + 4 = 0

⇒ 9x² + 4 × 1 = 0

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

9x² + 4(–i²) = 0

⇒ 9x² – 4i² = 0

⇒ (3x)² – (2i)² = 0

⇒ (3x + 2i)(3x – 2i) = 0 [∵ a² – b² = (a + b)(a – b)]

⇒ 3x + 2i = 0 or 3x – 2i = 0

⇒ 3x = –2i or 3x = 2i

Thus, the roots of the given equation are.

Given x² + 2x + 5 = 0

⇒ x² + 2x + 1 + 4 = 0

⇒ x² + 2(x)(1) + 1² + 4 = 0

⇒ (x + 1)² + 4 = 0 [∵ (a + b)² = a² + 2ab + b²]

⇒ (x + 1)² + 4 × 1 = 0

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

(x + 1)² + 4(–i²) = 0

⇒ (x + 1)² – 4i² = 0

⇒ (x + 1)² – (2i)² = 0

⇒ (x + 1 + 2i)(x + 1 – 2i) = 0 [∵ a² – b² = (a + b)(a – b)]

⇒ x + 1 + 2i = 0 or x + 1 – 2i = 0

⇒ x = –1 – 2i or x = –1 + 2i

∴ x = –1 ± 2i

Thus, the roots of the given equation are –1 ± 2i.

Given 4x² – 12x + 25 = 0

⇒ 4x² – 12x + 9 + 16 = 0

⇒ (2x)² – 2(2x)(3) + 3² + 16 = 0

⇒ (2x – 3)² + 16 = 0 [∵ (a + b)² = a² + 2ab + b²]

⇒ (2x – 3)² + 16 × 1 = 0

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

(2x – 3)² + 16(–i²) = 0

⇒ (2x – 3)² – 16i² = 0

⇒ (2x – 3)² – (4i)² = 0

Since a² – b² = (a + b)(a – b), we get

(2x – 3 + 4i)(2x – 3 – 4i) = 0

⇒ 2x – 3 + 4i = 0 or 2x – 3 – 4i = 0

⇒ 2x = 3 – 4i or 2x = 3 + 4i

Thus, the roots of the given equation are.

Given x² + x + 1 = 0

[∵ (a + b)² = a² + 2ab + b²]

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

[∵ a² – b² = (a + b)(a – b)]

Thus, the roots of the given equation are.

Given 4x² + 1 = 0

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

4x² – i² = 0

⇒ (2x)² – i² = 0

⇒ (2x + i)(2x – i) = 0 [∵ a² – b² = (a + b)(a – b)]

⇒ 2x + i = 0 or 2x – i = 0

⇒ 2x = –i or 2x = i

Thus, the roots of the given equation are.

Given x² – 4x + 7 = 0

⇒ x² – 4x + 4 + 3 = 0

⇒ x² – 2(x)(2) + 2² + 3 = 0

⇒ (x – 2)² + 3 = 0 [∵ (a – b)² = a² – 2ab + b²]

⇒ (x – 2)² + 3 × 1 = 0

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

(x – 2)² + 3(–i²) = 0

⇒ (x – 2)² – 3i² = 0

Since a² – b² = (a + b)(a – b), we get

Thus, the roots of the given equation are.

Given x² + 2x + 2 = 0

⇒ x² + 2x + 1 + 1 = 0

⇒ x² + 2(x)(1) + 1² + 1 = 0

⇒ (x + 1)² + 1 = 0 [∵ (a + b)² = a² + 2ab + b²]

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

(x + 1)² + (–i²) = 0

⇒ (x + 1)² – i² = 0

⇒ (x + 1)² – (i)² = 0

⇒ (x + 1 + i)(x + 1 – i) = 0 [∵ a² – b² = (a + b)(a – b)]

⇒ x + 1 + i = 0 or x + 1 – i = 0

⇒ x = –1 – i or x = –1 + i

∴ x = –1 ± i

Thus, the roots of the given equation are –1 ± i.

Given 5x² – 6x + 2 = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 5, b = –6 and c = 2

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given 21x² + 9x + 1 = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 21, b = 9 and c = 1

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given x² – x + 1 = 0

[∵ (a – b)² = a² – 2ab + b²]

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

[∵ a² – b² = (a + b)(a – b)]

Thus, the roots of the given equation are.

Given x² + x + 1 = 0

[∵ (a + b)² = a² + 2ab + b²]

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

[∵ a² – b² = (a + b)(a – b)]

Thus, the roots of the given equation are.

Given 17x² – 8x + 1 = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 17, b = –8 and c = 1

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given 27x² – 10x + 1 = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 27, b = –10 and c = 1

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given 17x² + 28x + 12 = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 17, b = 28 and c = 12

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given 21x² – 28x + 10 = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 21, b = –28 and c = 10

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given 8x² – 9x + 3 = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 8, b = –9 and c = 1

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given 13x² + 7x + 1 = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 13, b = 7 and c = 1

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given 2x² + x + 1 = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 2, b = 1 and c = 1

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here,, and

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here,, and

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here,, and

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here,, and

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here,, and

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

Given –x² + x – 2 = 0

⇒ x² – x + 2 = 0

[∵ (a – b)² = a² – 2ab + b²]

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

[∵ a² – b² = (a + b)(a – b)]

Thus, the roots of the given equation are.

Given

[∵ (a – b)² = a² – 2ab + b²]

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

[∵ a² – b² = (a + b)(a – b)]

Thus, the roots of the given equation are.

Given

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 3, b = –4 and

We have i² = –1

By substituting –1 = i² in the above equation, we get

Thus, the roots of the given equation are.

x² + 10ix – 21 = 0

Given x² + 10ix – 21 = 0

⇒ x² + 10ix – 21 × 1 = 0

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

x² + 10ix – 21(–i²) = 0

⇒ x² + 10ix + 21i² = 0

⇒ x² + 3ix + 7ix + 21i² = 0

⇒ x(x + 3i) + 7i(x + 3i) = 0

⇒ (x + 3i)(x + 7i) = 0

⇒ x + 3i = 0 or x + 7i = 0

∴ x = –3i or –7i

Thus, the roots of the given equation are –3i and –7i.

x² + (1 – 2i)x – 2i = 0

Given x² + (1 – 2i)x – 2i = 0

⇒ x² + x – 2ix – 2i = 0

⇒ x(x + 1) – 2i(x + 1) = 0

⇒ (x + 1)(x – 2i) = 0

⇒ x + 1 = 0 or x – 2i = 0

∴ x = –1 or 2i

Thus, the roots of the given equation are –1 and 2i.

Given

Thus, the roots of the given equation are and 3i.

6x² – 17ix – 12 = 0

Given 6x² – 17ix – 12 = 0

⇒ 6x² – 17ix – 12 × 1 = 0

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

6x² – 17ix – 12(–i²) = 0

⇒ 6x² – 17ix + 12i² = 0

⇒ 6x² – 9ix – 8ix + 12i² = 0

⇒ 3x(2x – 3i) – 4i(2x – 3i) = 0

⇒ (2x – 3i)(3x – 4i) = 0

⇒ 2x – 3i = 0 or 3x – 4i = 0

⇒ 2x = 3i or 3x = 4i

Thus, the roots of the given equation are and.

Given

Thus, the roots of the given equation are and 2i.

x² – (5 – i)x + (18 + i) = 0

Given x² – (5 – i)x + (18 + i) = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 1, b = –(5 – i) and c = (18 + i)

By substituting i² = –1 in the above equation, we get

By substituting –1 = i² in the above equation, we get

We can write 48 + 14i = 49 – 1 + 14i

⇒ 48 + 14i = 49 + i² + 14i [∵ i² = –1]

⇒ 48 + 14i = 7² + i² + 2(7)(i)

⇒ 48 + 14i = (7 + i)² [∵ (a + b)² = a² + b² + 2ab]

By using the result 48 + 14i = (7 + i)², we get

[∵ i² = –1]

∴ x = 2 + 3i or 3 – 4i

Thus, the roots of the given equation are 2 + 3i and 3 – 4i.

(2 + i)x² – (5 – i)x + 2(1 – i) = 0

Given (2 + i)x² – (5 – i)x + 2(1 – i) = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = (2 + i), b = –(5 – i) and c = 2(1 – i)

By substituting i² = –1 in the above equation, we get

We can write –2i = –2i + 1 – 1

⇒ –2i = –2i + 1 + i² [∵ i² = –1]

⇒ –2i = 1 – 2i + i²

⇒ –2i = 1² – 2(1)(i) + i²

⇒ –2i = (1 – i)² [∵ (a – b)² = a² – 2ab + b²]

By using the result –2i = (1 – i)², we get

[∵ i² = –1]

Thus, the roots of the given equation are 1 – i and.

x² – (2 + i)x – (1 – 7i) = 0

Given x² – (2 + i)x – (1 – 7i) = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 1, b = –(2 + i) and c = –(1 – 7i)

By substituting i² = –1 in the above equation, we get

We can write 7 – 24i = 16 – 9 – 24i

⇒ 7 – 24i = 16 + 9(–1) – 24i

⇒ 7 – 24i = 16 + 9i² – 24i [∵ i² = –1]

⇒ 7 – 24i = 4² + (3i)² – 2(4)(3i)

⇒ 7 – 24i = (4 – 3i)² [∵ (a – b)² = a² – b² + 2ab]

By using the result 7 – 24i = (4 – 3i)², we get

∴ x = 3 – i or –1 + 2i

Thus, the roots of the given equation are 3 – i and –1 + 2i.

ix² – 4x – 4i = 0

Given ix² – 4x – 4i = 0

⇒ ix² + 4x(–1) – 4i = 0

We have i² = –1

By substituting –1 = i² in the above equation, we get

ix² + 4xi² – 4i = 0

⇒ i(x² + 4ix – 4) = 0

⇒ x² + 4ix – 4 = 0

⇒ x² + 4ix + 4(–1) = 0

⇒ x² + 4ix + 4i² = 0 [∵ i² = –1]

⇒ x² + 2ix + 2ix + 4i² = 0

⇒ x(x + 2i) + 2i(x + 2i) = 0

⇒ (x + 2i)(x + 2i) = 0

⇒ (x + 2i)² = 0

⇒ x + 2i = 0

∴ x = –2i (double root)

Thus, the roots of the given equation are –2i and –2i.

x² + 4ix – 4 = 0

Given x² + 4ix – 4 = 0

⇒ x² + 4ix + 4(–1) = 0

We have i² = –1

By substituting –1 = i² in the above equation, we get

⇒ x² + 4ix + 4i² = 0

⇒ x² + 2ix + 2ix + 4i² = 0

⇒ x(x + 2i) + 2i(x + 2i) = 0

⇒ (x + 2i)(x + 2i) = 0

⇒ (x + 2i)² = 0

⇒ x + 2i = 0

∴ x = –2i (double root)

Thus, the roots of the given equation are –2i and –2i.

Given

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 2, and c = –i

By substituting i² = –1 in the above equation, we get

By substituting –1 = i² in the above equation, we get

We can write 15 – 8i = 16 – 1 – 8i

⇒ 15 – 8i = 16 + (–1) – 8i

⇒ 15 – 8i = 16 + i² – 8i [∵ i² = –1]

⇒ 15 – 8i = 4² + (i)² – 2(4)(i)

⇒ 15 – 8i = (4 – i)² [∵ (a – b)² = a² – b² + 2ab]

By using the result 15 – 8i = (4 – i)², we get

[∵ i² = –1]

Thus, the roots of the given equation are and.

x² – x + (1 + i) = 0

Given x² – x + (1 + i) = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 1, b = –1 and c = (1 + i)

By substituting –1 = i² in the above equation, we get

We can write 3 + 4i = 4 – 1 + 4i

⇒ 3 + 4i = 4 + i² + 4i [∵ i² = –1]

⇒ 3 + 4i = 2² + i² + 2(2)(i)

⇒ 3 + 4i = (2 + i)² [∵ (a + b)² = a² + b² + 2ab]

By using the result 3 + 4i = (2 + i)², we get

[∵ i² = –1]

∴ x = i or 1 – i

Thus, the roots of the given equation are i and 1 – i.

ix² – x + 12i = 0

Given ix² – x + 12i = 0

⇒ ix² + x(–1) + 12i = 0

We have i² = –1

By substituting –1 = i² in the above equation, we get

ix² + xi² + 12i = 0

⇒ i(x² + ix + 12) = 0

⇒ x² + ix + 12 = 0

⇒ x² + ix – 12(–1) = 0

⇒ x² + ix – 12i² = 0 [∵ i² = –1]

⇒ x² – 3ix + 4ix – 12i² = 0

⇒ x(x – 3i) + 4i(x – 3i) = 0

⇒ (x – 3i)(x + 4i) = 0

⇒ x – 3i = 0 or x + 4i = 0

∴ x = 3i or –4i

Thus, the roots of the given equation are 3i and –4i.

Given = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 1, and

By substituting i² = –1 in the above equation, we get

We can write

[∵ i² = –1]

[∵ (a – b)² = a² – 2ab + b²]

By using the result, we get

[∵ i² = –1]

Thus, the roots of the given equation are and.

xi.

Given

Thus, the roots of the given equation are and i.

2x² – (3 + 7i)x + (9i – 3) = 0

Given 2x² – (3 + 7i)x + (9i – 3) = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 2, b = –(3 + 7i) and c = (9i – 3)

By substituting i² = –1 in the above equation, we get

By substituting –1 = i² in the above equation, we get

We can write 16 + 30i = 25 – 9 + 30i

⇒ 16 + 30i = 25 + 9(–1) + 30i

⇒ 16 + 30i = 25 + 9i² + 30i [∵ i² = –1]

⇒ 16 + 30i = 5² + (3i)² + 2(5)(3i)

⇒ 16 + 30i = (5 + 3i)² [∵ (a + b)² = a² + b² + 2ab]

By using the result 16 + 30i = (5 + 3i)², we get

[∵ i² = –1]

Thus, the roots of the given equation are 3i and.

given (x-1)²+(x-2)²+(x-3)²=0

x² + 1 - 2x + X² + 4 - 4x + X² + 9 - 6x = 0

3X² - 12x+ 14 = 0

Comparing it with aX² + bx+ c = 0 and substituting them in b² – 4ac, we get

= (-12)²-4(3)(14)

= 144 – 168

= - 24 < 0 .

Hence the given equation do not have real roots. It has imaginary roots.

given x²- px + q = 0

We know sum of the roots = p

Product of the roots = q

As given that a and b are roots then,

a + b = p

a b = q

given

given α² + β² = 9

Given x²- px + 16 = 0 and α, β are roots of the equation then

Sum of roots α + β = p

Product of roots α β = 16

Substituting these in (α + β)²-2 α β=9 we get,

p² – 2(16) = 9

p² = 41

P = ±√41

we know irrational roots always exists in pair hence if is one root then is another root.

Given x² + px + q = 0

Sum of roots = -p

2+√3+ 2-√3 = -p

P = -4

Product of roots = q

(2+√3)(2-√3)=q

4 – 3 = q

q = 1.

given x² + ax + 8 = 0 and α – β = 2

Also from given equation α β = 8

As α – β = 2

Then

α² - 2 α – 8 =0

(α-4) (α + 2)=0

α = 4 and α = -2

if α = 4 then substituting it in α – β = 2 we get ,

β = 2

from the given equation,

sum of roots = -a

α + β = - a

- a = 4+2

a = -6

if α = - 2 then substituting it in α – β = 2 we get ,

β = - 4

then sum of roots α + β = - a

a = 6

therefore a = ± 6.

roots of a quadratic equation is

From the given equation we get,

Therefore

from the given equation sum of roots a + b = 1

Product of roots ab = 1

Now a² + b² = (a + b)²-2 a b

= 1 – 2

= - 1.

from the given condition roots remain unchanged only when they are equal to 1 and 0.

Hence the roots may be (0,1) or (1,0) and (1,1) and (0,0).

Hence 3 equations can be formed by substituting these points in (x-a) (x-b) = 0

Where a, b are roots or points.

from the given equation sum of the roots α + β = - l

Product of roots αβ = m

Formula to form a quadratic equation is x²- (α + β) x+ αβ = 0

Where α, β are roots of equation.

Given are roots, then required quadratic equation is

mx² – lx + 1 = 0.

given x²-a(x+1)-c=0

x²-ax-a-c=0

x²-ax-(a+c)=0

as α, β are roots of equation, we get

sum of the roots α + β = a

Product of roots αβ = - (a + c)

Given (1+ α) (1+ β)

= 1 + (α + β) + (α β)

= 1 + a – a – c

= 1 – c.

Since roots are equal then b² – 4ac = 0

From the given equation we get,

K² – 4 (1) (k+2) = 0

K² – 4 k - 8 = 0

=2±√12

given |x|²+|x|- 6 = 0

When x > 0

It can be written as x² + x – 6 = 0

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

X = 2

When x < 0

It can be written as x² - x – 6 = 0

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

X = - 2

Therefore x = ± 2

Hence sum of the roots = 0.

from the given equation sum of roots a + b = - 1

Product of roots ab = 1

Given a² + b²

= 1 – 2

= - 1.

given 4x²+ 3x + 7 = 0

We know sum of the roots =

Product of the roots =

As given that α and β are roots then,

given

given log₃(x² + 4x + 12) = 2

It can be written as

log₃(x² + 4x + 12) = 2log₃3

= log₃3²

log₃(x² + 4x + 12) = log₃9

x² + 4x + 12 = 9

x² + 4x + 3 = 0

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

X = -1, -3.

given (x²+2x)²-(x+1)²-55=0

[(x² +2x +1) -1] ² -(x +1)² -55 =0

(x +1)⁴ -2(x +1)² +1 -(x +1)² -55 =0

(x+1)⁴ -3(x +1)² -54 =0

let (x +1)²= r

r² -3r -54 =0

r² -9r +6r -54 =0

r( r -9) +6(r -9) =0

r = -6 , 9

but ( x+1)² ≥ 0 so, (x+1)² ≠ -6

so, (x +1)² = 9

x + 1 = ± 3

x = -1 ±3

x = -4, and 2 .

given

.

α² + pα + 1 = 0, β² + pβ + 1 = 0

α + β = -p, αβ = 1

γ² + qγ + 1 = 0, δ² + qδ + 1 = 0

δ – γ = , γδ = 1

(α – γ)(α + δ)(β – γ)(β + δ)

= (α² + α(δ – γ) – γδ)(β² + β(δ – γ) – δγ)

given |2x-x²-3| = 1

2x-x²-3 = ±1

When 2x-x²-3 = 1

⇒ 2x-x²-3 -1 = 0

⇒2x-x²-4 = 0

= x² – 2x +4 = 0

Discriminant, D = 4 - 16

= -12 < 0

Hence the roots are unreal.

When 2x-x²-3 = -1

= x² – 2x -2 = 0

Discriminant, D = 4 – 8 = - 4 < 0

Hence the roots are unreal.

Hence the given equation has no real roots.

when x > 0

x² + x-1 = 1

x² + x - 2 = 0

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

x = 1, -2

when x < 0

x² - x+1 = 1

x² – x = 0

x(x-1) = 0

x = 0, 1

hence the given equation has 3 solutions and they are x = 0, 1, -1.

(x² + x + 1)k = (x² – x + 1)

(k – 1)x² + (k + 1)x + (k – 1) = 0

For roots of quadratic equation real

Case I : a ≠ 0 and D ≥ 0

k – 1 ≠ 0 ⇒ k ≠ 1

-3k² + 10k – 3 ≥ 0

3k² – 10k + 3 ≤ 0

or

or

Case II : a = 0

k – 1 = 0 ⇒ k = 1

At k = 1, 2x = 0 ⇒ x = 0 is real

So, k = 1 is also count in answer.

Then, final answer is k ∈ [1/3, 3]

given that roots are consecutive, let they be a, a+1

From the formula for quadratic equation,

(x - a)(x - a - 1)
= x² - (a + 1)x - ax + a(a + 1)
= x² - (2a + 1)x + a(a + 1)
then
b² - 4c = (2a + 1)² - 4a(a + 1)
= 4a² + 1 + 4a - 4a² - 4a
= 1.

subtracting both the equations we get,

x² – 11x + a - x² +14x + 2a = 0

3x – a = 0

Substituting it in first equation we get,

a² – 24a = 0

a = 24.

given kx² + 1 = kx + 3x – 11x²

x²(k+11) – x(k + 3) + 1 = 0

as the roots are real and equal then the discriminant is equal to zero.

D = b² – 4ac = 0

(k+3)² – 4 (k+11) (1) = 0

K² + 9 + 6k – 4k – 44 = 0

K² + 2k – 35 = 0

(k-5) (k+7) = 0

K = 5, -7.

multiplying first equation and subtracting both the equations we get,

Substituting it in first equation we get,

given 4 is the root of x² + px + 12 = 0

= 16 + 4p + 12 = 0

4p = - 28

P = -7

Given x² + px + q = 0 has equal roots, then discriminant is 0.

D = b² – 4ac = 0

p² – 4q = 0

4q = 49

.

Sum of the roots = = -p

⇒ p + q = -p …(1)

Product of the roots = = q

⇒ pq = q

⇒ p = 1

Put value of p in eq.(1)

⇒ 1 + q = -1

⇒ q = -2

For roots to be real its D ≥ 0

(m + 1)² – 4(m + 4) ≥ 0

m² – 2m – 15 ≥ 0

(m – 1)² – 16 ≥ 0

(m – 1)² ≥ 16

m – 1 ≤ -4 or m – 1 ≥ 4

m ≤ -3 or m ≥ 5

For both roots to be negative product of roots should be

positive and sum of roots should be negative.

Product of roots = m + 4 > 0 ⇒ m > -4

Sum of roots = m + 1 < 0 ⇒ m < -1

After taking intersection of D ≥ 0, Product of roots > 0 and

sum of roots < 0. We can say that the final answer is

m ∈ (-4, -3]

given

(x+2) (x-5) (x+4) = (x-2) (x-3) (x+6)

x³+4x²-5x²-20x+2x²+8x-10x-40 = x³+6x²-3x²-18x-2x²-12x+6x+36

x²-22x-40 = x²-24x+36

4x = 76

x = 19

hence the given equation has only one solution.

given 4x²+ 3x + 7 = 0

We know sum of the roots =

Product of the roots =

As given that α and β are roots then,

given

from the given equation sum of the roots α + β = - p

Product of roots αβ = q

Formula to form a quadratic equation is x²- (α + β) x+ αβ = 0

Where α, β are roots of equation.

Given are roots, then required quadratic equation is

qx² – px + 1 = 0.

Difference of the roots

1

1

p² – 4q = 1

p² – 4q + 4q² – 4q² = 1

p² + 4q² = 1 + 2(2)(q) + (2q)²

p² + 4q² = (1 + 2q)²

given x²-p(x+1)-c=0

x²-px-p-c=0

x²-px-(p+c)=0

as α, β are roots of equation, we get

sum of the roots α + β = p

Product of roots αβ = - (p + c)

Given (1+ α) (1+ β)

= 1 + (α + β) + (α β)

= 1 + p – p – c

= 1 – c.

given that the equation has imaginary roots, hence the discriminant is less than 0.

= 25-4k<0

When we submit 7 in k the condition above will be satisfied and when we replace 6 the condition will be false.

So the least value of k is 7.

for the complex roots it will exists in pair.

Hence the roots are 1+i and 1-i

Formula for quadratic equation is (x-a) (x-b) = 0

(x-1-i) (x-1+i) = 0

x²-x+ix-x+1-i-ix+i-i²=0

x²-2x+2 = 0.