Question

If, pxx + qx + r = 0 ( p=0) quadratic equation's two roots are sin@ and cos@, then prove that (qq - pp) = 2rp​

Answer 1

Answer:

From the eq.,

product of roots = r/p

= > [email protected]@ = r/p

sum of roots = - q/p

= > sin@ + cos@ = -q/p

Square on both sides:

= > ( sin@ + cos@ )^2 = (-q/p)^2

= > sin^2@ + cos^2@ + 2sin@cos@ = q^2/p^2

= > 1 + 2( product of roots ) = q^2 / p^2

= > 1 + 2(r/p) = (q^2/p^2)

= > 2(r/p) = (q^2/p^2) - 1

= > (2r/p) = (q^2 - p^2 )/p^2

= > (2r/p) * p^2 = q^2 - p^2

= > 2rp = q^2 - p^2

= > q^2 - p^2 = 2rp

Proved.