If the roots of the equations (a2+b2)x2−2b(a+c)x+(b2+c2)=0 are equal, then


If the roots of the equations $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+\left(b^{2}+c^{2}\right)=0$ are equal, then

(a) $2 b=a+c$

(b) $b^{2}=a c$

(c) $b=\frac{2 a c}{a+c}$

(d) $b=a c$


The given quadric equation is $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+b^{2}+c^{2}=0$, and roots are equal.

Here, $a=\left(a^{2}+b^{2}\right), b=-2 b(a+c)$ and, $c=b^{2}+c^{2}$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=\left(a^{2}+b^{2}\right), b=-2 b(a+c)$ and, $c=b^{2}+c^{2}$

$=\{-2 b(a+c)\}^{2}-4 \times\left(a^{2}+b^{2}\right) \times\left(b^{2}+c^{2}\right)$

$=4 a^{2} b^{2}+4 b^{2} c^{2}+8 a b^{2} c-4\left(a^{2} b^{2}+a^{2} c^{2}+b^{4}+b^{2} c^{2}\right)$

$=+8 a b^{2} c-4 a^{2} c^{2}-4 b^{4}$

$=-4\left(a^{2} c^{2}+b^{4}-2 a b^{2} c\right)$

The given equation will have equal roots, if $D=0$

$-4\left(a^{2} c^{2}+b^{4}-2 a b^{2} c\right)=0$

$a^{2} c^{2}+b^{4}-2 a b^{2} c=0$

$\left(a c-b^{2}\right)^{2}=0$

$a c-b^{2}=0$

$a c=b^{2}$

Thus, the correct answer is (b)


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