Question

If roots of the equations $a{x}^2-2bx+c=0$ and $b{x}^2-2\sqrt{ac}x+b=0$ are real, then

• A. $ac=b^2$
• B. $4b^2-ac=0$
• C. $a=b,c=0$
• D. $a=b=0$

Answer 1

Answer: (A)

$ac=b^2$

Roots of the equations $a{x}^2-2bx+c=0$ and $b{x}^2-2\sqrt{ac}x+b=0$ are real.

$\Rightarrow 4b^2-4ac\ge 0$ and $4ac-4b^2\ge 0$

$\Rightarrow b^2-ac\ge 0$ and $b^2-ac\le 0$

$x\ge 0$ and $x\le 0$ iff $x=0$

$\therefore b^2=ac$

Hence, option A.