If $\alpha$ and $\beta$ are roots of $(x^2)-(4x)-1=0$, find $\sqrt[3]{\alpha}+ \sqrt[3]{\beta}$

Question

My question in handwriting https://i.sstatic.net/4vPBs.jpg

If $\alpha$ and $\beta$ are roots of this equation

$$(x^2)-(4x)-1=0$$

Then find $$\sqrt[3]{\alpha}+\sqrt[3]{\beta}$$

Please do not use the $\Delta = b^2-4ac$ method. Use Vieta's formulas: $$S=\alpha+\beta=-\frac{b}{a} \qquad P=\alpha\beta=\frac{c}{a}$$

Actually I want to solve the main entry using difference of squares! Such as (x-a)(x+a) And I note everybody that I know basical rules of quadratic equations like delta and etc.

Similar question here: https://math.stackexchange.com/q/1744538. — Martin R, Oct 11 '18 at 09:44

score 7 · Answer 1 · answered Oct 11 '18 at 09:37

7

Let $\sqrt[3]{\alpha}+\sqrt[3]{\beta}=x$.

Thus, since $\alpha+\beta=4$ and $\alpha\beta=-1,$ we obtain $$x^3=\alpha+\beta+3\sqrt[3]{\alpha\beta}(\sqrt[3]{\alpha}+\sqrt[3]{\beta})=4+3\cdot(-1)x$$ or $$x^3+3x-4=0$$ or $$x^3-x^2+x^2-x+4x-4=0$$ or $$(x-1)(x^2+x+4)=0,$$ which gives $x=1$.

answered Oct 11 '18 at 09:37

Michael Rozenberg

203,855

We get that $$\sqrt[3]{\alpha}+\sqrt[3]{\beta}$$ are NOT real numbers – Dr. Sonnhard Graubner Oct 11 '18 at 09:55
Why do you think so, Sonnhard? By the way, how are you? – Michael Rozenberg Oct 11 '18 at 09:56
I'm here in Austria for a few days – Dr. Sonnhard Graubner Oct 11 '18 at 09:59
Beautiful country! – Michael Rozenberg Oct 11 '18 at 10:00
Yes it is, my father was here in 1941 when he was 11 years old, because his mother has had an accident – Dr. Sonnhard Graubner Oct 11 '18 at 10:01
and by the way $$\sqrt[3]{2-\sqrt{5}}$$ is an complex number Michael – Dr. Sonnhard Graubner Oct 11 '18 at 10:04
Look at the link given by Martin R – Dr. Sonnhard Graubner Oct 11 '18 at 10:05
I don't agree with you! We can think about $\sqrt[3]{2-\sqrt5}$ like function $\sqrt[3]{x},$ which gives $\sqrt[3]{2-\sqrt5}=-\sqrt[3]{\sqrt5-2}.$ I saw the linked problem It's absolutely another problem because the given equation has no real roots. – Michael Rozenberg Oct 11 '18 at 10:06
1

Ok this is an old discussion about $$\sqrt[3]{x}$$! – Dr. Sonnhard Graubner Oct 11 '18 at 10:08

If $\alpha$ and $\beta$ are roots of $(x^2)-(4x)-1=0$, find $\sqrt[3]{\alpha}+ \sqrt[3]{\beta}$

1 Answers1