Question

If $a$ and $b$ are the non zero distinct roots of $x^2+ax+b=0$, then the minimum value of $x^2+ax+b$ is

• A. $\frac{2}{3}$
• B. $\frac{9}{4}$
• C. $-\frac{9}{4}$
• D. $-\frac{2}{3}$
• E. 1

Answer 1

Answer: (C)

Given, $x^2+ax+b=0$
For district now zero roots $D>0$
$\Rightarrow a^2-4b>0$
Now, $x^2+ax+b$
$={\left(x+\frac{a}{2}\right)}^2+\left(b-\frac{a^2}{4}\right)$
$={\left(x+\frac{a}{2}\right)}^2-\left(a^2-\frac{4b}{4}\right)$
We know, sum of roots $a+b=-a$
$\Rightarrow 2a+b=0$ ...(i)
Product of roots $a\times b=b$
$\Rightarrow b(a-1)=0$
$\Rightarrow a=1,b\ne 0$
From Eq. (i),
$2a+b=0$
$2(1)+b=0$
$b=-2$
Now, ${\left(x+\frac{a}{2}\right)}^2-\left(\frac{1^2+4\times 2}{4}\right)$
$={\left(x+\frac{a}{2}\right)}^2-\frac{9}{4}$
$\therefore$ Minimum value $=-\frac{9}{4}$