# For a,b,c non- zero, real distinct, the equation, ( a^2+b^2)x^2 -2b(a+c) x +b^2+c^2=0 has non-zero real roots. One of these roots is also the root of the equation? A) a^2x^2-a(b-c)x+bc=0 B) a^2x^2+a(c-b)x-bc=0 C) (b^2+c^2)x^2-2a(b+c)x+a^2=0

Mar 5, 2017

See below.

#### Explanation:

Solving for $x$ in $\left({a}^{2} + {b}^{2}\right) {x}^{2} - 2 b \left(a + c\right) x + {b}^{2} + {c}^{2} = 0$ we have

$x = \frac{b \left(a + c\right) - \sqrt{- {\left({b}^{2} - a c\right)}^{2}}}{{a}^{2} + {b}^{2}}$ The solutions must be real so necessarily ${b}^{2} - a c = 0$ and the roots are real identical

$x = \left\{\frac{c}{\sqrt{a c}} , \frac{c}{\sqrt{a c}}\right\}$

After substituting $b = \pm \sqrt{a c}$ into the equations from A) to C)

we obtain for the option B) the roots

$x = \left\{\frac{c}{\sqrt{a c}} , \frac{c}{a}\right\}$

so the correct option is B)