Question

Let $x^2-px+q=0$, where $pϵR,qϵR$ have roots $\alpha ,\beta$ such that $\alpha +2\beta =0$ then-

• A. $2p^2+q=0$
• B. $2q^2+p=0$
• C. $q<0$
• D. None of the above.

Answer 1

Answer: (A), (C)

The correct options are
A $2p^2+q=0$
C $q<0$

Given equation $x^2-px+q=0$ and roots of equation are α,β

sumation of roots α+β $=$ $\frac{-p}{1}=p$

and multiply of rootsαβ $=$ $\frac{q}{1}=q$

It is given that α+2β=0

$\Rightarrow$ (α+β)+β = 0

$\Rightarrow$ p+β = 0

$\Rightarrow$ β = $-p$

$\because$ 'β' is the root of equation $x^2-px+q=0$ so it will satisfy the equation

so, $(-p{)}^2-p(-p)+q=0$

$\Rightarrow$ $p^2+p^2+q=0$

$\Rightarrow$ $2p^2+q=0$

and $q=-2p^2$

$\because$ p,q both are real

so, $p^2$ will be positive number for all possible values of p

$\Rightarrow$ $p^2>0$

$\because$ $q=-2p^2$

so $q<0$

Hence the correct answers are

(A) $2p^2+q=0$

(C) $q<0$