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Question

# If the equation $\left({a}^{2}+{b}^{2}\right){x}^{2}-2b\left(a+c\right)x+\left({b}^{2}+{c}^{2}\right)=0$ has both roots equal, then (a) b = ac (b) $b=\frac{1}{2}\left(a+c\right)$ (c) b2 = ac (d) $b=\frac{2ac}{\left(a+c\right)}$

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Solution

## (c) b2 = ac $\text{It is given that the roots of the equation}\left\{\left({a}^{2}+{b}^{2}\right){x}^{2}-2b\left(a+c\right)x+\left({b}^{2}+{c}^{2}\right)=0\right\}\text{are equal}\text{.}\phantom{\rule{0ex}{0ex}}\therefore \left({b}^{2}-4ac\right)=0\phantom{\rule{0ex}{0ex}}⇒{\left\{2b\left(a+c\right)\right\}}^{2}-4\left({a}^{2}+{b}^{2}\right)\left({b}^{2}+{c}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒4{b}^{2}\left({a}^{2}+2ac+{c}^{2}\right)-4\left({a}^{2}{b}^{2}+{a}^{2}{c}^{2}+{b}^{4}+{b}^{2}{c}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒4{a}^{2}{b}^{2}+8a{b}^{2}c+4{b}^{2}{c}^{2}-4{a}^{2}{b}^{2}-4{a}^{2}{c}^{2}-4{b}^{4}-4{b}^{2}{c}^{2}=0\phantom{\rule{0ex}{0ex}}⇒8a{b}^{2}c-4{a}^{2}{c}^{2}-4{b}^{4}=0\phantom{\rule{0ex}{0ex}}⇒2a{b}^{2}c-{a}^{2}{c}^{2}-{b}^{4}=0\phantom{\rule{0ex}{0ex}}⇒{b}^{4}-2ac{b}^{2}+{a}^{2}{c}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{\left({b}^{2}-ac\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}-ac=0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=ac\phantom{\rule{0ex}{0ex}}$

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