Roots of $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0$ will be real only if a,b,c are in G.P. and in this case these roots will be equal as well.

True

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False

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Solution

The correct option is A True
(a) $B^{2} - 4 A C \geq 0$ for roots to be real.
Now $B^{2} - 4 A C = 4 b^{2} {(a + c)}^{2} - 4 (a^{2} + b^{2}) (b^{2} + c^{2})$
$= 4 [b^{2} a^{2} + b^{2} c^{2} + 2 b^{2} a c - (a^{2} b^{2} + a^{2} c^{2} + b^{4} + b^{2} c^{2})]$
$= 4 [2 b^{2} a c - a^{2} c^{2} - b^{4}]$
$= - 4 [b^{4} + a^{2} c^{2} - 2 b^{2} a c]$
$= - 4 {[b^{2} - a c]}^{2}$ .
Now $= - 4 {[b^{2} - a c]}^{2}$ is greater than or equal to zero, and hence the discriminant $B^{2} - 4 A C$ is always $\leq 0$ so that the roots cannot be real unless $b^{2} - a c = 0$ or $b^{2} = a c$ i.e. a,b,c are in G.P. In this case the discriminant being zero the roots will be equal also.

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