Question

Solve for $x:4x^2-4ax+\left(a^2{b}^2\right)=0$
OR
Solve for x : $3x^2-2\sqrt{6x}+2=0$

Answer 1

The given quadratic equation is $4x^2-4ax+(a^2-b^2)=0$
$4x^2-4ax+(a^2-b^2)=0$
$\therefore 4x^2-4ax+(a-b)(a+b)=0$
$\Rightarrow 4x^2+[-2a-2a+2b-2b]x+(a-b)(a+b)=0$
$\Rightarrow 4x^2+(2b-2a)x-2(2a+2b)x+(a-b)(a+b)=0$
$\Rightarrow 4x^2+2(b-a)x-2(a+b)x+(a-b)(a+b)=0$
$\Rightarrow 2x[2x-(a-b\left)\right]-(a+b)[2x-(a-b\left)\right]=0$
$\Rightarrow [2x-(a-b\left)\right]$ or $[2x-(a+b\left)\right]=0$
$\Rightarrow 2x=a-b$ or 2x =a+b
$\Rightarrow x=\frac{a-b}{2}$ or $x=\frac{a+b}{2}$
Thus, the solution of the given quadratic equation is given by $x=\frac{a-b}{2}$ or $x=\frac{a+b}{2}$
OR
The given quadratic equation is $3x^2-2\sqrt{6x}+2=0$
Comparing with the quadratic equation $a{x}^2+bx+c=0$, we have
$a=3,b=-2\sqrt{6}$ and c =2
discriminant of the given quadratic equation
$D=b^2-4ac=(2\sqrt{6}{)}^2-4\times 3\times 2=24-24=0$
$\therefore x=\frac{-2\left(\sqrt{6}\right)\pm \sqrt{0}}{6}\because x=\frac{-b\pm \sqrt{D}}{2a}$
$\Rightarrow x=\frac{2\sqrt{6}}{6}\Rightarrow x=\frac{\sqrt{6}}{3}$
Thus, the solution of the given quadratic equation is $x=\frac{\sqrt{6}}{3}$