Quadratic equations (roots real)

Question

Complete set of real values of $a$ for which the equation $x^4-2ax^2+x+a^2-a=0$ has all its roots real?

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My attempt:

I tried to make a quadratic in $x$ so that i could use $D>0$ condition but it doesn't lead me to values of $a$

Answer 1

As it turns out, your expression can be factorized: $$ x^4-2ax^2 + x + a^2-a = (a-x^2-x) (a-x^2+x-1) $$

Once this is done, you can write down the roots of your polynomial explicitly using the quadratic formula: $$ x = \frac{1}{2} (- 1 \pm \sqrt{4a+1}) \\ x = \frac{1}{2} ( 1 \pm \sqrt{4a-3}) $$

Now, all roots are real if and only if the square roots above are of positive numbers, so that complex roots don't come up because of a negative square root. This is equivalent to $4a+1 \geq 0$ and $4a-3 \geq 0$. The first happens when $a\geq -0.25$ and the second when $a \geq 0.75$.

Hence, both happen together when $a \geq 0.75 = \frac 34$. Hence all the roots are real when $a \geq \frac 34$.

Answer 2

It's better to see it as an equation in $a$. You have $a^2 - a (2x^2 +1) + x^4 + x=0$ with discriminant $(2x -1)^2 \ge 0, \forall x$. Therefore $a_{1,2} = \frac {2x^2 + 1 \pm (2x -1)} 2$. From here: $(a - x^2 -x)(a - x^2 + x -1)=0$

Should be easy from here.