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

If a and b are the non-zero distinct roots of x2+ax+b=0x^2 + ax + b = 0, then the least value x2+ax+bx^2 + ax + b is

A

32\frac{3}{2}

B

94\frac{9}{4}

C

−94-\frac{9}{4}

D

1

Answer

-\frac{9}{4}

Explanation

Solution

Given that aa and bb are the roots of

x2+ax+b=0,x^2 + ax + b = 0,

by Viète’s formulas:

a+b=−a⟹b=−2a,a + b = -a \quad \Longrightarrow \quad b = -2a, ab=b⟹a(−2a)=−2a⟹−2a2=−2a.ab = b \quad \Longrightarrow \quad a(-2a) = -2a \quad \Longrightarrow \quad -2a^2 = -2a.

Since a≠0a \neq 0, dividing both sides by −2a-2a gives:

a=1.a = 1.

Thus,

b=−2.b = -2.

The quadratic becomes:

x2+x−2.x^2 + x - 2.

The least (minimum) value of a quadratic f(x)=x2+x−2f(x) = x^2 + x - 2 occurs at x=−12x = -\frac{1}{2} (vertex formula x=−b2ax = -\frac{b}{2a}):

f(−12)=(−12)2+(−12)−2=14−12−2=1−2−84=−94.f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 2 = \frac{1}{4} - \frac{1}{2} - 2 = \frac{1 - 2 - 8}{4} = -\frac{9}{4}.