Inequality with discriminants

Question

If $x^2-ax+1-2a^2>0$ for all $x \in {R}$, find range of $a$

The solution to this takes the discriminant of the expression in terms of $a$, i.e., $$\implies D={a^2-4(1-2a)}>0(\because x \in R)$$ and then figures out the range from the inequality in $a$.

But what I dont understand is, as per my knowledge, the discriminant is taken from a quadratic equation, so in order for $D$ to be as it is in the solution, we must first consider $$x^2-ax+1-2a^2=0$$

What I mean to say is,

Say $x^2-ax+1-2a^2=y$

$$\implies x^2-ax+1-2a^2-y=0$$ $$D=a^2-4(1-2a^2-y)>0$$

But the solution has considered only $a^2-4(1-2a^2-y)>0$, which means that they have considered $y=0$, but the question says that $y>0$.

The question is clearly stating that the expression is greater than and not equal to $0$. So how can we work with the discriminant here?

Answer 1

There are already a few solutions provided, but I believe I can offer a succinct explanation for the pre-calculus level. This is how I am taught in the Singapore O-level additional mathematics syllabus

Recall that for a quadratic equation $ax^2+bx+c=0$, descriminant = $b^2-4ac$

when discriminant $> 0$, the equation has 2 real and distinct roots

when discriminant $= 0$, the equation has 2 real roots that are of the same magnitude

when discriminant $<0$, the equation has 2 complex roots

Given $x^2−ax+1−2a^2>0$, $x^2−ax+1−2a^2=0$ has no real solutions. Why? Well, if the expression given is always more than $0$, then it will never equal $0$, so it cannot have any roots.

Since $x^2−ax+1−2a^2=0$ has no real solutions, so discriminant $<0$, or

$$a^2-4(1)(1-2a^2)<0$$ $$a^2-4+8a^2<0$$ $$9a^2-4<0$$ $$(3a-2)(3a+2)<0$$ Sketching the graph, we get $-2/3<a<2/3$

Answer 2

Maybe the question is the following: find all $a$'s such that the inequality $x^2-ax+1-2a^2>0$ holds for all $x\in \mathbb{R}$.

Now you look at the polynomial in $x$ above ( $a$ is a parameter) and you impose the condition that the discriminant is negative. That gives $$D_a=9 a^2 - 4 <0$$

Answer 3

For very (very, very) large $|x|$, $$x^2-ax+1-2a^2\approx x^2>0$$ This may be hard to see — how large $|x|$ needs to be depends on $a$. But for any fixed value of $a$ there is some point beyond which the $x^2$ term dominates.

What about for small $|x|$? In particular, choose $x$ to make $x^2-ax+1-2a^2$ as small as possible, and call the chosen $x$-value $x_0$. There are two possibilities:

• If $$x_0^2-ax_0+1-2a^2\geq0$$ then for every other value of $x$ $$x^2-ax+1-2a^2\geq x_0^2-ax_0+1-2a^2\geq0$$
• If $x_0^2-ax_0+1-2a^2<0$, then as $x$ ranges between $x_0$ and very large (or very small) values, there is some intermediate point where $$x^2-ax+1-2a^2=0$$

Thus the inequality in the problem fails iff $x^2-ax+1-2a^2=0$ has at two solutions — one between $-\infty$ and $x_0$ and one between $x_0$ and $\infty$. The discriminant helps you answer that equivalent question.