If the roots of the equation $\left(a^{2}+b^{2}\right) x^{2}-2(a c+b d) x+\left(c^{2}+d^{2}\right)=0$ are equal, prove that $\frac{a}{b}=\frac{c}{d}$.
The given quadric equation is $\left(a^{2}+b^{2}\right) x^{2}-2(a c+b d) x+\left(c^{2}+d^{2}\right)=0$, and roots are real
Then prove that $\frac{a}{b}=\frac{c}{d}$.
Here,
$a=\left(a^{2}+b^{2}\right), b=-2(a c+b d)$ and, $c=\left(c^{2}+d^{2}\right)$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=\left(a^{2}+b^{2}\right), b=-2(a c+b d)$ and, $c=\left(c^{2}+d^{2}\right)$
$D=b^{2}-4 a c$
$=\{-2(a c+b d)\}^{2}-4 \times\left(a^{2}+b^{2}\right) \times\left(c^{2}+d^{2}\right)$
$=4\left(a^{2} c^{2}+2 a b c d+b^{2} d^{2}\right)-4\left(a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2}\right)$
$=-4 a^{2} d^{2}-4 b^{2} c^{2}+8 a b c d$
$=-4\left(a^{2} d^{2}+b^{2} c^{2}-2 a b c d\right)$
The given equation will have real roots, if $D=0$
$-4\left(a^{2} d^{2}+b^{2} c^{2}-2 a b c d\right)=0$
$\left(a^{2} d^{2}+b^{2} c^{2}-2 a b c d\right)=0$
$(a d)^{2}+(b c)^{2}-2(a d)(b c)=0$
$(a d-b c)^{2}=0$
Square root both sides we get,
$a d-b c=0$
$a d=b c$
$\frac{a}{b}=\frac{c}{d}$
Hence $\frac{a}{b}=\frac{c}{d}$
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