Find $m$ so that given equation has real roots

Question

We are given the following equation:

$$x^4 - (2m - 1)x^2 + 4m - 5 = 0, m \in \mathbb{R}$$

Find all $m$'s so that the given equation has real roots.

I thought I only had to put $\Delta \geq 0 $. However, the official solution adds two more conditions: $S \geq 0$ and $P \geq 0$, where $S$ and $P$ are the sum and the product of the roots.

Can someone explain why do we need the last two conditions?

Thank you in advance!

Answer 1

Yes. To solve this biquadratic equation, you set $y=x^2$ and solve first the quadratic equation $$y^2-(2m-1)y+4m-5=0,$$ whence the condition $\Delta\ge 0$, but also you must have non-negative roots, since they're squares.

This equivalent to $P\ge 0$ (roots have the same sign). When they have the same sign, this sign is also the sign of their sum, so $S\ge0$.

Answer 2

Let $y=x^2$. Then we have the quadratic:

$$y^2−(2m−1)y+4m−5=0$$

This quadratic in $y$ has real root(s) if and only if

$$(2m-1)^2-4(4m-5)=(2m-7)(2m-3)\geq 0$$

But we also need that the root(s) of the quadratic in $y$ are positive (so that $y=x^2$ has real solutions)