Finding a quadratic equation with roots $x=α^2$ and $x=β^2$

Question

I'm struggling on a question and have no idea where to start.

The quadratic equation $y=ax^2+bx+c$ has roots $x=\alpha$ and $x=\beta$. Find an equation that has roots $x=\alpha^2$ and $x=\beta^2$.

I was just wondering if anyone could provide any knowledge so I can start this question?

First of all, there's a lot to clarify about your question. Who is $a$? Who is $d$? Is the quadratic really $\alpha x^2+bx+c$ ? — LHF, Mar 09 '20 at 22:40
I made a mistake in the title, there is no d, sorry! But yes, that's the quadratic. a isn't specified either. — user123942345, Mar 09 '20 at 22:44

Oscar Lanzi · Answer 1 · 2020-03-10T20:17:47.407

2

Apply the Graeffe root-squaring method.

$ax^2+bx+c=0$

$ax^2+c=-bx$ separating even and odd degree terms

$a^2x^4+2acx^2+c^2=b^2x^2$ after squaring

$\color{blue}{a^2(x^2)^2+(2ac-b^2)(x^2)+c^2=0}$

edited Mar 10 '20 at 20:17

answered Mar 10 '20 at 00:51

Oscar Lanzi

score 1 · Accepted Answer · answered Mar 09 '20 at 22:48

1

Assuming you want to express the quadratic with roots $\alpha^2$ and $\beta^2$ in terms of $a,b,c$, from Vieta's:

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\frac{b^2}{a^2}-\frac{2c}{a}$$

$$\alpha^2\beta^2=\frac{c^2}{a^2}$$

Can you end it from here?

answered Mar 09 '20 at 22:48

LHF

I wasn't sure if I had to use Vieta's formula or not, thanks! I've got it now. – user123942345 Mar 09 '20 at 22:53

score 1 · Answer 3 · answered Mar 09 '20 at 22:48

Hints:

Can you express the sum $\alpha+\beta$ and the product $\alpha\beta$ of the original roots in terms of the original coefficients $a,b,c$?
Can you express the sum $\alpha^2+\beta^2$ and the product $\alpha^2\beta^2$ of the new roots in terms of the sum $\alpha+\beta$ and the product $\alpha\beta$ of the original roots?
Can you express the new coefficients $a',b',c'$ in terms of the sum $\alpha^2+\beta^2$ and the product $\alpha^2\beta^2$ of the new roots?

3 Answers3