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Question

# (i) If the roots of the equation $\left({a}^{2}+{b}^{2}\right){x}^{2}-2\left(ac+bd\right)x+\left({c}^{2}+{d}^{2}\right)=0$ are equal, prove that $\frac{a}{b}=\frac{c}{d}$. [CBSE 2017] (ii) If ad ≠ bc then prove that the equation $\left({a}^{2}+{b}^{2}\right){x}^{2}+2\left(ac+bd\right)x+\left({c}^{2}+{d}^{2}\right)=0$ has no real roots. [CBSE 2017]

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Solution

## (i) It is given that the roots of the equation $\left({a}^{2}+{b}^{2}\right){x}^{2}-2\left(ac+bd\right)x+\left({c}^{2}+{d}^{2}\right)=0$ are equal. $\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{\left[-2\left(ac+bd\right)\right]}^{2}-4\left({a}^{2}+{b}^{2}\right)\left({c}^{2}+{d}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒4\left({a}^{2}{c}^{2}+{b}^{2}{d}^{2}+2abcd\right)-4\left({a}^{2}{c}^{2}+{a}^{2}{d}^{2}+{b}^{2}{c}^{2}+{b}^{2}{d}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒4\left({a}^{2}{c}^{2}+{b}^{2}{d}^{2}+2abcd-{a}^{2}{c}^{2}-{a}^{2}{d}^{2}-{b}^{2}{c}^{2}-{b}^{2}{d}^{2}\right)=0$ $⇒\left(-{a}^{2}{d}^{2}+2abcd-{b}^{2}{c}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒-\left({a}^{2}{d}^{2}-2abcd+{b}^{2}{c}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒{\left(ad-bc\right)}^{2}=0\phantom{\rule{0ex}{0ex}}$ $⇒ad-bc=0\phantom{\rule{0ex}{0ex}}⇒ad=bc\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\frac{c}{d}$ Hence Proved. (ii) Consider the equation $\left({a}^{2}+{b}^{2}\right){x}^{2}+2\left(ac+bd\right)x+\left({c}^{2}+{d}^{2}\right)=0$ On comparing the above equation with the general equation $a{x}^{2}+bx+c$, we get $a=\left({a}^{2}+{b}^{2}\right),b=2\left(ac+bd\right),c=\left({c}^{2}+{d}^{2}\right)$ We know that $D={b}^{2}-4ac$ $⇒D={\left\{2\left(ac+bd\right)\right\}}^{2}-4×\left({a}^{2}+{b}^{2}\right)×\left({c}^{2}+{d}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒D=\left\{4{\left(ac+bd\right)}^{2}\right\}-4×\left({a}^{2}+{b}^{2}\right)×\left({c}^{2}+{d}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒D=4\left\{{a}^{2}{c}^{2}+{b}^{2}{d}^{2}+2abcd\right\}-4\left\{{a}^{2}{c}^{2}+{a}^{2}{d}^{2}+{b}^{2}{c}^{2}+{b}^{2}{d}^{2}\right\}\phantom{\rule{0ex}{0ex}}⇒D=4\left\{{a}^{2}{c}^{2}+{b}^{2}{d}^{2}+2abcd-{a}^{2}{c}^{2}-{a}^{2}{d}^{2}-{b}^{2}{c}^{2}-{b}^{2}{d}^{2}\right\}\phantom{\rule{0ex}{0ex}}⇒D=4\left\{-{a}^{2}{d}^{2}-{b}^{2}{c}^{2}+2abcd\right\}\phantom{\rule{0ex}{0ex}}⇒D=-4{\left(ad-bc\right)}^{2}$ Since, the value of D is negative, the given equation has no real roots.

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