Question

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

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

Answer 1

Answer: (B)

As given a and b are the roots of the equation
$x^2+ax+b=0$
$\Rightarrow$ sum of roots, a + b = - a $\Rightarrow b=-2a$ ...(1)
and product of roots, ab = b
$\Rightarrow ab-b=0$
$\Rightarrow b(a-1)=0$
if b = 0 then a = 0
if b $\ne$ 0 then a = 1 and b = - 2
so, the expression will be,
$f(x)=x^2+x-2$
$=x^2+2.\frac{1}{2}x+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2-2$
$\Rightarrow f\left(x\right)={\left(x+\frac{1}{2}\right)}^2-\frac{9}{4}$
So, f (x) will be minimum, if ${\left(x+\frac{1}{2}\right)}^2=0$
i.e. when $x=-\frac{1}{2}$
$\Rightarrow$ minimum value of function $=-\frac{9}{4}$