Let $a \in \mathbb R$. Prove that ($x^2 + ax + a > 0$ for every $x \in \mathbb R)$ iff $(0< a<4)$

Question

1. List item

This is a homework question for my university math proofs course. I am asked to prove the above bi conditional statement. The text book gives the hint that completing the square of the quadratic equation will simplify it.

Thanks, John

Answer 1

If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:

$$x^2+ax+a=(x-\frac a2)^2-\frac{a^2}{4}+a$$

The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $a\in(0,4)$, as stated in the question.

Answer 2

Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.

Answer 3

Let $x^2+ax+a=y$

$\iff x^2+ax+a-y=0$

As $x$ is real, the discriminant must be $\ge0$

$a^2-4(a-y)\ge0\iff4y\ge a(4-a)$

It is sufficient to have $a(4-a)>0\iff a(a-4)<0\iff0<a<4$

Answer 4

Complete the square: $$x^2+ax+a>0 \iff \left(x+\frac a2\right)^2-\frac{a^2}{4}+a>0 \iff \\ \left(x+\frac a2\right)^2>\frac{a^2-4a}{4} \iff \left(x+\frac a2\right)^2\ge 0>\frac{a^2-4a}{4}, 0<a<4.$$

Answer 5

After completing the square of the above equation you will get: $$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$ which is nothing but: $$(x+\frac{a}{2})^2 + (a-\frac{a^2}{4})$$ The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+\frac{a}{2})^2\gt0$$ Now for the second part $$a-\frac{a^2}{4}\gt0$$ which gives $$4a-a^2\gt0$$ Or $$a^2-4a\lt0$$ $$a(a-4)\lt0$$ which gives the critical points $a=0,4$ Now this inequality only holds if $0\lt a\lt4$

Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers. One more thing is you can check for the vertex of the parabola.... Hope this helps