If $a$ and $b$ are the non-zero distinct roots of the equation $x^2+ax+b=0$, find the least value of $x^2+ax+b$

Question

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

My Approach:
The way I have adopted is quite long. Solving
$f(a)=0$
$f(b)=0$
we obtain $a^4-a^3-a^2=0$
The roots of this equation are $0,0,\frac{1+ \root\of{5}}{2},\frac{ 1- \root\of{5}}{2}$. Now it is given that $a,b\neq 0$. Hence the first two roots are eliminated. On calculating the values for b we obtain $b=-(3+\root\of5), (-3+\root\of5)$.
Now once we get the 2 values of $a,b$ we put them in the equation and calculate the minimum value of the function for both cases. Comparing the 2 minimum values obtained we get the answer.

Is there an easier or shorter way of solving this question as this solution seems too long for an MCQ...

Answer 1

Given $a$ and $b$ are roots of the equation $x^2+ax+b=0.$

Sum of the roots: $a+b=-a.........(1)$
Product of the roots $ab=b.......(2)$ $$(1)\Rightarrow 2a+b=0 \text { and } (2)\Rightarrow a=1 \text { or }b=0$$ Thus, $a=1$ and $b=-2$.

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So, we have $x^2+x-2=(x+\frac {1}{2})^2-\frac {9}{4}=0.$
Hence minimum value of $x^2+ax+b=0$ is $-\frac {9}{4} $. Hope it helps.

Answer 2

Comparing $$x^2 + ax +b = (x-a)(x-b)$$ one gets $a=1$ and $b=-2$. Now one needs to find the minimum of $$x^2 + x -2$$ which is of course $-\dfrac 94$.