Find all pairs $(a,b)$ of real numbers such that $\alpha$ is a root of $x^2 +ax + b =0$, $\alpha^2 -2$ is also a root of the equation.

Question

I have tried to solve the question above by taking $2$ cases.

CASE: $1$
Let the 2 roots of the equation be $\alpha$ and $\beta$. In this case I took $\alpha = \beta$ and I got 2 solutions for $(a,b)$.

CASE: $2$
In this case I took $\alpha$ is not equal to $\beta$.
But after this I am not able to solve it .

Answer 1

Hint:

We must have: $$x^2+ax+b=(x-\alpha)(x-\alpha^2+2)$$

Working out the RHS and comparing it with LHS we find expressions for $a$ and $b$ in $\alpha$.

Answer 2

Since $(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$

Sum $s$ and product $p$ of the roots $\{x_1,x_2\}$ verify the special quadratic equation $x^2-sx+p=0$

Thus $\begin{cases}a=-\alpha-(\alpha^2-2)\\b=\alpha(\alpha^2-2)\end{cases}$

And this is done. You don't need to have subcases (checking if the roots are equal or distinct [see comment]), you just need to express $a$ and $b$ in function of the given data, namely $\alpha$.

Edit:

In the case $\alpha=\alpha^2-1\iff \alpha\in\{-1,2\}$ then we can also consider that $(x-\alpha)(x-\mu)=0$ is also satisfying the wording of the problem for any $\mu$, while $\alpha$ is not necessarily a double root of this quadratic.

So, in this case, there is an additional infinite family of solutions which is

$\begin{cases}a=-\alpha-\mu\\b=\alpha\mu\end{cases}\quad \forall \mu\in\mathbb R$

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Note that in both cases $(a,b)$ belongs to the line $a\alpha+b=-\alpha^2$:

• it is the whole line when $\alpha=-1\text{ or }2$
• else it is just a point on it

Bonus, here is the locus of $(a,b)$ when $\alpha$ describe $\mathbb R$.

Answer 3

By solving the quadratic equation,

$$\alpha=\frac{-a+\sqrt{a^2-4b}}{2}$$ is a root.

Then for the other root,

$$\frac{-a-\sqrt{a^2-4b}}{2}=\left(\frac{-a+\sqrt{b^2-4b}}{2}\right)^2-2.$$

And symmetrically, we have the solutions

$$\frac{-a+\sqrt{a^2-4b}}{2}=\left(\frac{-a-\sqrt{a^2-4b}}{2}\right)^2-2.$$

This forms the locus of $(a,b)$.