Question

Solve the following quadratic equation by completing the square: $4x^2+4bx-(a^2-b^2)=0$

• A. $\{\frac{a-2}{2},-\left(\frac{a+2}{2}\right)\}$
• B. $\{\frac{a+2}{2},-\left(\frac{a+2}{2}\right)\}$
• C. $\{\frac{a-2}{2},-\left(\frac{a-2}{2}\right)\}$
• D. None of these

Answer 1

The correct option is D None of these

Given equation is $4x^2+4bx-(a^2-b^2)=0$

$\Rightarrow$ $x^2+bx-\frac{(a^2-b^2)}{4}=0$

$\Rightarrow$ $x^2+bx=\frac{(a^2-b^2)}{4}$

Now, ${\left(\frac{b}{2}\right)}^2=\frac{b^2}{4}$

Adding $\frac{b^2}{4}$ on bothe sides,

$\Rightarrow$ $x^2+bx+\frac{b^2}{4}=\frac{a^2}{4}-\frac{b^2}{4}+\frac{b^2}{4}$

$\Rightarrow$ ${(x+\frac{b}{2})}^2=\frac{a^2}{4}$

Taking square root on both sides

$\Rightarrow$ $x+\frac{b}{2}=\pm \frac{a}{2}$

$\Rightarrow$ $x+\frac{b}{2}=\frac{a}{2}$ and $x+\frac{b}{2}=-\frac{a}{2}$

$\Rightarrow$ $x=\frac{a-b}{2}$ and $x=-\left(\frac{a+b}{2}\right)$