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

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

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Answer 1 · 2017-03-05T14:51:29

See below.

Explanation:

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

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

$x = {\frac{c}{\sqrt{a c}}, \frac{c}{\sqrt{a c}}}$ $x = {c/sqrt(ac),c/sqrt(ac)}$

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

we obtain for the option B) the roots

$x = {\frac{c}{\sqrt{a c}}, \frac{c}{a}}$ $x = {c/sqrt(ac),c/a}$

so the correct option is B)

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