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The roots of the quadratic equation $2 x^{2} + ax - a^{2} = 0$ are [[1]] and [[2]].

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answer is A 2, - A.

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Detailed Solution

The roots of the quadratic equation

2 x^{2} + ax - a^{2} = 0

are

\frac{a}{2}

and

- a

.
Given quadratic equation is,

2 x^{2} + ax - a^{2} = 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

2 x^{2} + ax - a^{2} = 0

with standard form we get,

a = 2, b = a, c = - a^{2}

Then,

D = b^{2} - 4 ac

⟹ D = a^{2} - 4 \times 2 \times (- a^{2})

⟹ D = a^{2} + 8 a^{2}

⟹ D = 9 a^{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{- a \pm \sqrt{9 a^{2}}}{2 \times 2}

⟹ x = \frac{- a \pm 3 a}{4}

On taking positive sign,

⟹ x = \frac{- a + 3 a}{4}

⟹ x = \frac{2 a}{4}

⟹ x = \frac{a}{2}

On taking negative sign,

⟹ x = \frac{- a - 3 a}{4}

⟹ x = \frac{- 4 a}{4}

⟹ x = - a

Therefore, the roots of the equation

2 x^{2} + ax - a^{2} = 0

are

\frac{a}{2}

and

- a

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