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The roots of the equation $x^{2} + 5 x - (a^{2} + a - 6) = 0$ are [[1]] and [[2]].

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answer is ( A - 2 ), - A + 3.

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Detailed Solution

The roots of the equation

x^{2} + 5 x - (a^{2} + a - 6) = 0

are

a - 2

and

- (a + 3)

.
Given quadratic equation is,

x^{2} + 5 x - (a^{2} + a - 6) = 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} + 5 x - (a^{2} + a - 6) = 0

with standard form we get,

a = 1, b = 5, c = - (a^{2} + a - 6)

Then,

D = b^{2} - 4 ac

⟹ D = 5^{2} - 4 \times 1 \times \{- (a^{2} + a - 6)\}

⟹ D = 25 + 4 a^{2} + 4 a - 24

⟹ D = {4 a}^{2} + 4 a + 1

⟹ D = {(2 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{- 5 \pm \sqrt{{(2 a + 1)}^{2}}}{2 \times 1}

⟹ x = \frac{- 5 \pm (2 a + 1)}{2}

On taking positive sign,

⟹ x = \frac{- 5 + (2 a + 1)}{2}

⟹ x = \frac{- 5 + 2 a + 1}{2}

⟹ x = \frac{- 4 + 2 a}{2}

⟹ x = a - 2

On taking negative sign,

⟹ x = \frac{- 5 - (2 a + 1)}{2}

⟹ x = \frac{- 5 - 2 a - 1}{2}

⟹ x = \frac{- 6 - 2 a}{2}

⟹ x = - (a + 3)

Therefore, the roots of the equation

x^{2} + 5 x - (a^{2} + a - 6) = 0

are

a - 2

and

- (a + 3)

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