Quadratic Inequality

Algebra Level 3

$x^{2} + ax+a^{2} + 6a <0$

Find the sum of squares of all integral values of $a$ for which the inequality above is satisfied for all $1<x<2.$

The answer is 91.

2 solutions

Prakhar Gupta
Mar 13, 2015

Let us write:- $f(x) = x^{2} + ax + a^{2} + 6a$ We claim that if and only if both $f(1)$ and $f(2)$ are non-positive, then the given inequality is satisfied.

Solving first inequality:- $f(1) \le 0$ $a^{2} +7a + 1\le 0$ This implies (approximately) that $a \in (-6.85,-0.146)$ .

Solving second inequality:- $f(2) \le 0$ $a^{2} + 8a + 4 \le 0$ This implies (approximately) that $a \in (-7.46, -0.53)$ .

Integers which satisfies both inequalities are $\{-1,-2,-3,-4,-5,-6\},$ so the answer is $1^2+2^2 + \ldots + 6^2 = 91.$

Good work. You mentioned that the inequality is satisfied if and only if $f(1),f(2)\le 0.$ Can someone explain why this is the case?

Eli Ross Staff - 5 years, 7 months ago

if x > 0 then f(x) ≥ f(y) if x ≥ y so if f(1) and f(2) are both ≤ 0 then f(x) ≤ 0 for 1 ≤ x ≤ 2.

Lu Ca - 1 year, 11 months ago

Can someone explain why we need $f(1), f(2)$ to be non-positive?

Anik Mandal - 4 years, 8 months ago

Karan Shekhawat
Mar 8, 2015

ans is ${(-1)}^{2}+{(-2)}^{2}+{(-3)}^{2}+{(-4)}^{2}+{(-5)}^{2}+{(-6)}^{2}$

Please explain how... It would be really grateful of u =)

Yogesh Verma - 6 years, 3 months ago

Quadratic Inequality

The answer is 91.

2 solutions

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