Question

Solve the following quadratic equation for x:

x2 - 4ax - b2 + 4a2 = 0

Answer 1

x^2-4ax+4a^2-b^2=0
x^2-2*2a*x+(2a)^2-(b)^2=0
(x-2a)^2-(b)^2=0
(x-2a-b)(x-2a+b)=0
x-2a-b=0        x-2a+b=0
x=b+2a          x=2a-b
Therefore, x= 2a+b or 2a-b

Answer 2

$2a+b$ and $2a-b$ are the solutions of the equation $x^2-4ax-b^2+4a^2=0$.

Given:

An equation $x^2-4ax-b^2+4a^2=0$.

To Find:

The solution of the equation.

Solution:

Rewrite the given equation.

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

Write $4ax=2\cdot 2a\cdot x$ and $4a^2=(2a)^2$ in the above equation.

$(x^2-2\cdot 2a\cdot x+(2a)^2)-b^2=0$

The expression $x^2-2\cdot 2a\cdot x+(2a)^2$ can be written as $(x^2-2\cdot 2a\cdot x+(2a)^2)=(x-2a)^2$ because it is the same as the identity $(a-b)^2=a^2-2ab+b^2$. Rewrite the equation using the mentioned identity.

$(x-2a)^2-b^2=0$

Now the obtained equation follows the identity $a^2-b^2=(a-b)(a+b)$. Use the mentioned identity and simplify to find $x$.

$(x-2a-b)(x-2a+b)=0\\x=2a+b\, , \,x=2a-b$

There are two values of $x$ are obtained here.

Thus, $2a+b$ and $2a-b$ are the solutions of the equation $x^2-4ax-b^2+4a^2=0$.