Question

If the equation $a{x}^2+bx+6=0$ does not have two distinct real roots, then the least value of $3a+b$ is

• A. $3$
• B. $-3$
• C. $2$
• D. $-2$

Answer 1

Answer: (D)

$-2$

As the quadratic equation does no have two distinct real roots, so $b^2=4ac$

$b^2=4a\times 6=24a$

$b^2=24a$

$3a=\frac{b^2}{8}$

$3a+b=\frac{b^2}{8}+b=\frac{b^2+8b}{8}=\frac{(b+4{)}^2-16}{8b}$

Least value of $(b+4{)}^2=0$

Hence least value of $\frac{(b+4{)}^2-16}{8b}=\frac{0-16}{8}=-2$