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

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Detailed Solution

The roots of the equation

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

are

\frac{a^{2} + b^{2}}{2}

and

\frac{a^{2} - b^{2}}{2}

.
Given quadratic equations is,

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

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

with standard form we get,

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

Then,

D = b^{2} - 4 ac

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

⟹ D = 16 a^{4} - 16 a^{4} + 16 b^{4}

⟹ D = 16 b^{4} > 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}^{2}) \pm \sqrt{16 b^{4}}}{2 \times 4}

⟹ x = \frac{{4 a}^{2} \pm {4 b}^{2}}{8}

⟹ x = \frac{a^{2} \pm b^{2}}{2}

On taking positive sign,

⟹ x = \frac{a^{2} + b^{2}}{2}

On taking negative sign,

⟹ x = \frac{a^{2} - b^{2}}{2}

Therefore, the roots of the equation

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

are

\frac{a^{2} + b^{2}}{2}

and

\frac{a^{2} - b^{2}}{2}

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