Question:
Show that the equation $2\left(a^{2}+b^{2}\right) x^{2}+2(a+b) x+1=0$ has no real roots, when $a \neq b$.
Solution:
The quadric equation is $2\left(a^{2}+b^{2}\right) x^{2}+2(a+b) x+1=0$
Here,
$a=2\left(a^{2}+b^{2}\right), b=2(a+b)$ and,$c=1$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=2\left(a^{2}+b^{2}\right), b=2(a+b)$ and, $c=1$
$D=\{2(a+b)\}^{2}-4 \times 2\left(a^{2}+b^{2}\right) \times 1$
$=4\left(a^{2}+2 a b+b^{2}\right)-8\left(a^{2}+b^{2}\right)$
$=4 a^{2}+8 a b+4 b^{2}-8 a^{2}-8 b^{2}$
$=8 a b-4 a^{2}-4 b^{2}$
$D=-4\left(a^{2}-2 a b+b^{2}\right)$
$=-4(a-b)^{2}$
We have,
Thus, the value of $D<0$
Therefore, the roots of the given equation are not real
Hence, proved