Question

If $a$ and $b$ $(\ne 0)$ are the roots of the equation $x^2+ax+b=0$, then find the least value of $x^2+ax+b$ $(x\in R)$.

• A. $\frac{9}{4}$
• B. $\frac{1}{4}$
• C. $-\frac{9}{4}$
• D. $-\frac{1}{4}$

Answer 1

The correct option is C $-\frac{9}{4}$

Since $a$ and $b$ are the roots of equation $x^2+ax+b=0$
We have,
$a+b=-a$, $ab=b$
Now,
$ab=b$ or $(a-1)b=0$ or $a=1$
Substitute $a=1$ in $a+b=-a$, we get $b=-2$.
Hence, $x^2+ax+b=x^2+x-2=(x+\frac{1}{2}{)}^2-\frac{1}{4}-2$
$=(x+\frac{1}{2}{)}^2-\frac{9}{4}$
Which has a minimum value equal to $-\frac{9}{4}$.