Question

If one root of the quadratic equation x-px +q = 0 be double of the other then find
the relation between p and q.​

Answer 1

Given :

• One root of the quadratic equation x² - px + q = 0 is the double the other root

To Find :

• The relation between p and q

Solution :

Let one of the root be a , then other root becomes 2a {given condition}

Now , the relation applying the relation between roots and coefficients of the quadratic equation ,

• Sum of the roots of the quadratic equation is equal to the coefficient of x
• Product of the roots of quadratic equation is equal to the constant

Applying the first condition ,

➙ 2a + a = -p

➙ 3a = -p

➙ $\sf{a = \dfrac{-p}{3}}$ .........(1)

Applying the second condition ,

➙2a(a) = q

➙ 2a² = q

➙ a² = $\sf{\dfrac{q}{2}}$

Substituting the value of a here {i.e , from eq(1) }

➙ $\sf{\dfrac{q}{2} ={\dfrac{-p}{3}}^2}$

➙ $\sf{\dfrac{q}{2} = \dfrac{p^2}{9}}$

By Cross multiplication ,

➙ ${\boxed{\underline{\bf{\pink{2p^2=9q}}}}}$

Hence ,

• The relation between p and q is 2p² = 9q

Answer 2

Given:-

One root of the quadratic equation x-px +q = 0 be double of the other

To Find:-

relation between p and q

Solution:-

Let ß be one root and 2ß be the 2nd root.

$\implies \sf x^{2}-px +q = 0$

The sum of the roots=P

ß+2ß=P

ß=$\sf\dfrac{p}{3}$---------(1)

The product of the roots =q

2ß.ß=q

$\sf 2ß ^{2}=q$

$\sf 2(\dfrac{p}{3})^{2}=q$-----(from1)

$\sf\green{2p^2=9q}$

The relation between p and q

$\sf\blue{\boxed{2p^2=9q}}$