Question

If the equation $a{x}^2+bx+6=0$ has real roots , where $a,b\in R$, then the greatest value of $3a+b$,is

• A. $4$
• B. $-1$
• C. $-2$
• D. $1$

Answer 1

Answer: (C)

$-2$

Given equation has real roots $\Rightarrow b^2>24a$

Finding minimum value of $3a+b$

As $a<\left(b^2/24\right)$ $3a+b<3\left(\frac{b^2}{24}\right)+b$

Let $f\left(b\right)=\frac{b^2}{8}+b$

For finding we will differentiate the equation $\Rightarrow f^{{}^{\prime }}\left(b\right)=2b/8=-1\Rightarrow b=-4$

$\Rightarrow a=b^2/24\Rightarrow a=2/3$

Minimum value of $3a+b=3\left(2/3\right)-4=-2$

$\therefore$minimum value =$-2$