Question

Prove that x²(p²-q²)+2x(pr-qs)+(r²-s²)=0 has real roots.

Answer 1

Answer:

p, q, r, s should be real numbers

Step-by-step explanation:

$D/4=b'^2-ac$ is commonly used for finding the nature of the root.

$D/4=(pr-qs)^2-(p^2-q^2)(r^2-s^2)$

$D/4=p^2r^2-2pqrs+q^2s^2-(p^2r^2-p^2s^2-q^2r^2+q^2s^2)$

$D/4=p^2r^2-2pqrs+q^2s^2-p^2r^2+p^2s^2+q^2r^2-q^2s^2$

$D/4=-2pqrs+p^2s^2+q^2r^2$

$D/4=(ps-qr)^2\geq 0$ for real number p, q, r, s

∴$D/4\geq 0$, true for real number p, q, r, s