Solve using quadratic formula : 4x² + 4ax + (a² + b²).

Solution:

A quadratic equation is in the form of ax² + bx + c = 0. 

Let's find value of x that satisfies 4x² + 4ax + (a² + b²) = 0.

Given, 4x² + 4ax + (a² + b²) = 0

Roots are given by the formula: x = (- b ±√ b² - 4ac) / 2a

As we know that coefficient of x²  is a, coefficient of x is b and constant term is c, so, a = 4, b = 4a and c = (a² + b²)

Using the quadratic formula, we get,

⇒ - 4 a ± √(4a)² - 4 (4) (a² + b²) / 2 (4) 

⇒ - 4 a ± √(16 a²- 16 ( a² + b²) / 8

⇒ x = - 4 a ± √(16 a² - 16 a² -16 b²) / 8

⇒ x = - 4 a ± √(-16b²)/ 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 4x² + 4ax + (a² + b²).

Solve using quadratic formula : 4x² + 4ax + (a² + b²).

Summary:

x  = (-a ± bi)/2 are the solutions for equation 4x² + 4ax + (a² + b²)