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]

Answer 1

(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.
$$\begin{gathered} \therefore D=0 \\ \Rightarrow {\left[-2\left(ac+bd\right)\right]}^2-4\left(a^2+b^2\right)\left(c^2+d^2\right)=0 \\ \Rightarrow 4\left(a^2{c}^2+b^2{d}^2+2abcd\right)-4\left(a^2{c}^2+a^2{d}^2+b^2{c}^2+b^2{d}^2\right)=0 \\ \Rightarrow 4\left(a^2{c}^2+b^2{d}^2+2abcd-a^2{c}^2-a^2{d}^2-b^2{c}^2-b^2{d}^2\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(-a^2{d}^2+2abcd-b^2{c}^2\right)=0 \\ \Rightarrow -\left(a^2{d}^2-2abcd+b^2{c}^2\right)=0 \\ \Rightarrow {\left(ad-bc\right)}^2=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow ad-bc=0 \\ \Rightarrow ad=bc \\ \Rightarrow \frac{a}{b}=\frac{c}{d} \end{gathered}$$
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
$a=(a^2+b^2),b=2(ac+bd),c=(c^2+d^2)$
We know that $D=b^2-4ac$
$$\begin{gathered} \Rightarrow D={\left\{2\left(ac+bd\right)\right\}}^2-4\times \left(a^2+b^2\right)\times \left(c^2+d^2\right) \\ \Rightarrow D=\left\{4{\left(ac+bd\right)}^2\right\}-4\times \left(a^2+b^2\right)\times \left(c^2+d^2\right) \\ \Rightarrow D=4\left\{a^2{c}^2+b^2{d}^2+2abcd\right\}-4\left\{a^2{c}^2+a^2{d}^2+b^2{c}^2+b^2{d}^2\right\} \\ \Rightarrow D=4\left\{a^2{c}^2+b^2{d}^2+2abcd-a^2{c}^2-a^2{d}^2-b^2{c}^2-b^2{d}^2\right\} \\ \Rightarrow D=4\left\{-a^2{d}^2-b^2{c}^2+2abcd\right\} \\ \Rightarrow D=-4{\left(ad-bc\right)}^2 \end{gathered}$$
Since, the value of D is negative, the given equation has no real roots.