Show that the equation 2(a2+b2)x2+2(a+b)x+1=0 has no real roots,

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