Question

The condition that the equation $x^2+px+q=0$, whose one root is the cube of the other root is :

• A. $p=q^{1/4}[1-q^{1/2}]$
• B. $-p=q^{1/2}[1-q^{1/4}]$
• C. $-p=q^{1/4}[1+q^{1/2}]$
• D. $p=q^{1/2}[1+q^{1/4}]$

Answer 1

The correct option is C $-p=q^{1/4}[1+q^{1/2}]$

The given equation is

$x^2+px+q=0$

If $\alpha$ and $\beta$ are the roots of the quadratic equation $A{x}^2+Bx+C=0$, then

$\alpha +\beta =-\frac{B}{A}$

$\alpha \beta =\frac{C}{A}$

for given equation,

$\alpha +\beta =-p$........(1)

$\alpha \beta =q$............(2)

Now, it is given that

$\alpha ={\beta }^3$.........(3)

from equation (2) and (3),

${\beta }^4=q$

$\beta =q^{\frac{1}{4}}$

and

$\alpha ={\beta }^3$

$=q^{\frac{3}{4}}$

now, from (1),

$\alpha +\beta =-p$

$q^{\frac{3}{4}}+q^{\frac{1}{4}}=-p$

$q^{\frac{1}{4}}[q^{\frac{1}{2}}+1]=-p$