Question

If the roots of the equation $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$ are equal then =

• A. $2b=ac$
• B. $b^2=ac$
• C. $b=\frac{2ac}{a+c}$
• D. $b=ac$

Answer 1

The correct option is B $b^2=ac$

$(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$
Roots are real and equal $\therefore$ $D=0$

$D=b^2-4ac=0$
$\Rightarrow [-2b(a+c){]}^2-4(a^2+b^2\left)\right(b^2+c^2)=0$
$\Rightarrow b^2(a^2+c^2+2ac)-(a^2{b}^2+a^2{c}^2+b^4+c^2{c}^2)=0$
$\Rightarrow b^2{a}^2+b^2{c}^2+2ac{b}^2-a^2{b}^2-a^2{c}^2-b^4-b^2{c}^2=0$
$\Rightarrow 2ac{b}^2-a^2{c}^2-2ac{b}^2=0$
$\Rightarrow (b^2-ac{)}^2=0$
$\Rightarrow b^2=ac$