Find the domain of the inequation $x^2 -ax +1-2a^2 > 0$ for all $x\in \Bbb R$

Question

if $x^2 -ax +1-2a^2 > 0$ for all $x\in \Bbb R$ then find the set of real values of $a$.

the book give the solution as :

since $x^2 -ax +1-2a^2 > 0$ for all $x\in \Bbb R$, $\quad \implies \quad a^2-4(1-2a^2) < 0\quad \implies \quad -\frac{2}3<a<\frac{2}3$

but is the discriminant is smaller than $0$, then the solutions skips into the complex region. then how can the value of $a$ be such that the solutions of the inequation are complex.

Answer 1

actuelly we get $$x^2-2\cdot \frac{a}{2}x+\frac{a^2}{4}+1-2a^2-\frac{a^2}{4}>0$$ this is $$\left(x-\frac{a}{2}\right)^2+1-\frac{9}{4}a^2>0$$

Answer 2

$y_a:= x^2 - ax+ 1-2a^2$ can be thought of a parabola, which opens upward .

Completing the square gives you the vertex.

1) $y_a \gt 0$ for all $x$, the parabola does not have zeroes, i.e. does not touch or go below the $X$-axis.

$y_a$ has no zeroes: $y= x^2-ax+1-2a^2 =0$ has no real roots.

This is the case for:

$a^2 -4(1-2a^2) \lt 0$, or

$9a^2 -4 \lt 0$, or

$(3a -2)(3a+2) \lt 0.$

One factor must be positive the other negative.

Hence : $-2/3 < a< 2/3.$

Answer 3

but if the discriminant is smaller than 0, then the solutions skips into the complex region. then how can the value of a be such that the solutions of the inequation are complex.

That is precisely what you WANT.

If the equation had real solutions then sometimes $x^2 -ax +1-2a^2 = 0$. But you want it to NEVER equal $0$. So you want it to have NO real solutions.

Okay.... that does mean it has complex solutions but you don't care about those. You only care that $x^2 -ax +1-2a^2 > 0$.

And for that to happen the discriminant MUST be negative. That is what you WANT. If the discriminant were $0$ or greater then there would be real solutions which you absolutely do NOT want to ever happen.