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The roots of the equation $x^{2} - 4 ax + (4 a^{2} - b^{2}) = 0$ are [[1]] and [[2]].

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answer is 2 A + B, 2 A - B.

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Detailed Solution

The roots of the equation

x^{2} - 4 ax + (4 a^{2} - b^{2}) = 0

are

2 a + b

and

2 a - b

.
Given quadratic equation is,

x^{2} - 4 ax + (4 a^{2} - b^{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

x^{2} - 4 ax + (4 a^{2} - b^{2}) = 0

with standard form we get,

a = 1, b = - 4 a, c = (4 a^{2} - b^{2})

Then,

D = b^{2} - 4 ac

⟹ D = {(- 4 a)}^{2} - 4 \times 1 \times (4 a^{2} - b^{2})

⟹ D = {16 a}^{2} - {16 a}^{2} + {4 b}^{2}

⟹ D = {4 b}^{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{- (- 4 a) \pm \sqrt{{4 b}^{2}}}{2 \times 1}

⟹ x = \frac{4 a \pm 2 b}{2}

⟹ x = 2 a \pm b

On taking positive sign,

⟹ x = 2 a + b

On taking negative sign,

⟹ x = 2 a - b

Therefore, the roots of the equation

x^{2} - 4 ax + (4 a^{2} - b^{2}) = 0

are

2 a + b

and

2 a - b

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