Question

Let $\alpha ,\beta$ be the roots of $x^2-ax+b=0$, where $a$ & $b\in$ $R$. If $\alpha +3\beta =0$, then

• A. $3a^2+4b=0$
• B. $3b^2+4a=0$
• C. $b<0$
• D. $a<0$

Answer 1

Answer: (A), (C)

The correct options are
A $3a^2+4b=0$
C $b<0$

$\alpha ,\beta$ are the roots of $x^2-ax+b=0$ and $\alpha +3\beta =0$

i.e. $\alpha +\beta =a$ and $\alpha \beta =b$

$\alpha +3\beta =(\alpha +\beta )+2\beta =a+2\beta =0$

$\Rightarrow \beta =\frac{-a}{2}$ and $\alpha =\frac{3a}{2}$

Substituting $\beta$ value in the equation gives

$\frac{a^2}{4}+\frac{a^2}{2}+b=0$

$\therefore 3a^2+4b=0$

$b=\alpha \beta =\frac{3a}{2}\left(\frac{-a}{2}\right)=\frac{-3a^2}{4}$

$\Rightarrow b<0$

Hence, options A and C.