Question

If $x^2+3x+3=0$ and $a{x}^2+bx+1=0$, $a,b,\in ,Q$ have a common root, then value of $(3a+b)$ is equal to

• A. $\frac{1}{3}$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

Q If $x^2+3x+3=0$ and $a{x}^2+bx+1=0$ & a,b $ϵ$ Q have

a common root then volume (3a+b) is equal to

$\to$ Let p,q be the common root between $x^2+3x+3=0$

and $a{x}^2+bx+1=0$

we know from properties of quadratic equation that

$\to$ sum of roots , $p+q=\frac{-b}{a}=-3$

$\therefore b=3a$

$\to$ product of root $,pq=\frac{c}{a}=\frac{1}{a}=3$

$\therefore a=1/3$

$\therefore b=3\times \frac{1}{3}=1$

$\therefore$ volume of $3a+b=3\times \frac{1}{3}+1$

3a+b=2