Question

prove the given equation$x^2(p^2+q^2)+2x(pr+qs)+r^2+s^2=0$ has no real roots. If $ps=qr,$ , then show that the roots are real and equal.

Answer 1

The given Quadratic equation is,

$x^2(p^2+q^2)+2x(pr+qs)+r^2+s^2=0$

Here, $a=p^2+q^2,b=2(pr+qs) ,c=r^2+s^2$

Now,

Δ$= b^2-4ac=[2(pr+qs)]^2-4(p^2+q^2)(r^2+s^2)$

Δ$= b^2-4ac=4[p^2r^2+2pqrs+q^2s^2-p^2r^2-p^2s^2-q^2r^2-q^2s^2]$

Δ $= b^2-4ac=4[-p^2s^2+2pqrs-q^2r^2]$

Δ$= -4[(ps-qr)^2]<0\: ...(1)$

Since, Δ $= b^2-4ac<0,$ the roots are not real.

If $ps=qr$ then Δ= $-4[ps-qr]^2 = -4[qr-qr]^2<0 \:\:\:$  (using(1))

Thus, Δ = 0 if $ps=qr$ and so the roots will be real and equal.