# Solve using quadratic formula : 4x^{2} + 4ax + (a^{2} + b^{2}).

**Solution:**

A quadratic equation is in the form of ax^{2 }+ bx + c = 0.

Let's find value of x that satisfies 4x^{2} + 4ax + (a^{2} + b^{2}) = 0.

Given, 4x² + 4ax + (a² + b²) = 0

Roots are given by the formula: x = (- b ±√ b^{2} - 4ac) / 2a

As we know that coefficient of x^{2 } is a, coefficient of x is b and constant term is c, so, a = 4, b = 4a and c = (a^{2 }+ b^{2})

Using the quadratic formula, we get,

⇒ - 4 a ± √(4a)^{2 }- 4 (4) (a^{2 }+ b^{2}) / 2 (4)

⇒ - 4 a ± √(16 a^{2}- 16 ( a^{2 }+ b^{2}) / 8

⇒ x = - 4 a ± √(16 a^{2 }- 16 a^{2 }-16 b^{2}) / 8

⇒ x = - 4 a ± √(-16b^{2})/ 8

⇒ x = - 4 a ± (-4bi)/ 8

x = 4 (- a ± bi ) / 8

We can have two values of 'x' that satisfies the equation.

⇒ x =(- a + bi) / 2 or x = (- a - bi) / 2

⇒x = (-a ± bi)/2

Thus, x =(-a ± bi)/2 are the solutions for equation 4x^{2} + 4ax + (a^{2} + b^{2}).

## Solve using quadratic formula : 4x^{2} + 4ax + (a^{2} + b^{2}).

**Summary:**

x = (-a ± bi)/2 are the solutions for equation 4x^{2} + 4ax + (a^{2} + b^{2})

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