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The roots of the equation $x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ are [[1]] and [[2]].

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answer is A - 2, - A + 4.

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Detailed Solution

The roots of the equation

x^{2} + 6 x - (a^{2} + 2 a - 8) = 0

are

a - 2

and

- (a + 4)

.
Given quadratic equation is,

x^{2} + 6 x - (a^{2} + 2 a - 8) = 0

.
We know that,
Discriminant for the quadratic equation

a x^{2} + bx + c = 0, a \neq 0

where a, b and c are real numbers is given by,

D = b^{2} - 4 ac

.
On comparing the given equation

x^{2} + 6 x - (a^{2} + 2 a - 8) = 0

with standard form we get,

a = 1, b = 6, c = - (a^{2} + 2 a - 8)

Then,

D = b^{2} - 4 ac

⟹ D = 6^{2} - 4 \times 1 \times \{- (a^{2} + 2 a - 8)\}

⟹ D = 36 + {4 a}^{2} - 8 a - 32

⟹ D = {4 a}^{2} - 8 a + 4

⟹ D = {4 (a + 1)}^{2} > 0

Discriminant is greater than zero which means the roots of the equation exist.
Since, quadratic formula for the quadratic equation

a x^{2} + bx + c = 0, a \neq 0

where a, b and c are real numbers is given by,

x = \frac{- b \pm \sqrt{D}}{2 a}

Then,

x = \frac{- 6 \pm \sqrt{{4 (a + 1)}^{2}}}{2 \times 1}

⟹ x = \frac{- 6 \pm 2 (a + 1)}{2}

⟹ x = - 3 \pm (a + 1)

On taking positive sign,

⟹ x = - 3 + (a + 1)

⟹ x = a - 2

On taking negative sign,

⟹ x = - 3 - (a + 1)

⟹ x = - a - 4 = - (a + 4)

Therefore, the roots of the equation

x^{2} + 6 x - (a^{2} + 2 a - 8) = 0

are

a - 2

and

- (a + 4)

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