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

2 b = a c

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b^{2} = a c

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b = \frac{2 a c}{a + c}

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b = ac

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Solution

The correct option is A $b^{2} = a c$
Since the roots are equal
$B^{2} - 4 A C = 0$
Hence
$4 b^{2} (a + c)^{2} - 4 (b^{2} + c^{2}) (a^{2} + b^{2}) = 0$
$4 b^{2} a^{2} + 4 b^{2} c^{2} + 8 b^{2} a c - 4 b^{2} a^{2} - 4 b^{4} - 4 c^{2} a^{2} - 4 c^{2} b^{2} = 0$
$8 b^{2} a c - 4 b^{4} - 4 c^{2} a^{2} = 0$
Or
$(b^{2})^{2} - 2 b^{2} a c + (a c)^{2} = 0$
$b^{2} = \frac{2 a c \pm \sqrt{4 a c^{2} - 4 a c^{2}}}{2}$
$b^{2} = a c$

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