Question

If one root of the equation x2-px+q=0 be double of the other, show that 2p2=9q

Answer 1

Answer:

Given: $x^{2} -px + q = 0$; One of the roots is twice the other

Past knowledge: We know that for a quadratic equation of the form $ax^{2} +bx+c=0$

Sum of its roots = $\frac{-b}{a}$

Product of its roots = $\frac{c}{a}$

To Prove: $2p^{2} = 9q$

Proof:

Let $\alpha$ and $\beta$ be the roots of the above equation

The question states that one root is twice the other.

⇒ $\alpha = 2\beta$

Sum of the roots of the equation = $\alpha +\beta = \frac{-(-p)}{1} = p$            (Past Knowledge)

⇒ $\alpha +\beta = p$

We know that $\alpha =2\beta$                                                                 (Given)

⇒ $2\beta + \beta = p$

⇒ $3\beta = p$                                                                                      

⇒ $\beta = \frac{p}{3}$                                                                                       $(1)$

Product of roots of the equation = $\alpha \beta = \frac{q}{1} = q$

⇒ $\alpha \beta =q$

We know that $\alpha =2\beta$                                                                 (Given)

⇒ $(2\beta )\beta = q$

⇒ $2\beta ^{2} = q$

We know that $\beta = \frac{p}{3}$                                                                  $(1)$

⇒ $2(\frac{p}{3})^{2} = q$

⇒ $2(\frac{p^{2} }{9} )= q$

⇒ $2p^{2} = 9q$

$Hence$ $Proved$