Question

On solving $4x^2-4a^{2}x+(a^4-b^4)$ = 0 we get value of $x$ is equal to

• A. $\frac{a^2\pm b^2}{2}$
• B. $\frac{a^2}{2}$
• C. $\frac{b^2}{2}$
• D. $\frac{a^2÷b^2}{2}$

Answer 1

The correct option is A $\frac{a^2\pm b^2}{2}$

On comparing the equation with $A{x}^2+Bx+C=0$,
We get, $A=4,B=-4a^2,C=a^4-b^4$
D = $B^2-4AC$ =$(-4a^2{)}^2$ $-$ 4*4*$(a^4-b^4)$
= $16a^4-16a^4+16b^4$= $16b^4$
$x=\frac{-B\pm \sqrt{D}}{2A}$ = $\frac{-4(-4a^2)\pm \sqrt{16b^4}}{2\times 4}$
= $\frac{4a^2\pm 4b^2}{8}$ = $\frac{a^2+b^2}{2}$ or $\frac{a^2-b^2}{2}$