An algebra problem by Satvik Golechha

Algebra Level 3

Determine the product of all the possible real values of the expression $\large \sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ .

Details: You may look up Vieta's Formula

The answer is 1.

1 solution

Satvik Golechha
Jul 16, 2014

Let $\large x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$

Whole-cubing both sides, we get upon simplification-

$\large x^3+3x-4=0$ , So the product of all roots is $4$ ......GOSH!!

The expression is real, and so can have only real values. We see that $\large x=1$ satisfies the equation. On dividing by $\large x-1$ , we get $\large x^2+x+4$ , which has no real roots..... so answer is $\large \boxed 1$

@Satvik Golechha You ask for the product of all roots, both real and imaginary. Since the polynomial is of degree 3, there will be 3 roots, and by Descartes' Rule of Signs we know that there will be 1 (positive) real root and 2 complex roots.

I like your approach using Vieta's to find the product of the roots, but you have applied the formula incorrectly. The product of all the roots will actually equal $(-1)^{3} * \frac{-4}{1} = 4$ , and not $-4$ . Your answer of 1 doesn't make sense, given that you explicitly allowed for both real and imaginary roots.

After first entering 4 as the answer and being told that it was wrong, I spent quite some time figuring out my mistake. I don't think I did make a mistake; I strongly believe that the answer is 4, and that you should adjust the "official" solution accordingly.

Nevertheless, you have posted a good question. :) Regards.

Brian Charlesworth - 6 years, 11 months ago

Sorry for that small error....that minus sign was unintentional... But I think 1 is correct, the real expression cannot take imaginary values...

Satvik Golechha - 6 years, 11 months ago

I think that the answer would be $64$ because you are asking for the product of all possible values of that expression. We see that there are two cube roots, and every cube root has three possible values, so in this case we have $9$ possible values.

Let $a=\sqrt[3]{2+\sqrt{5}}$ and $b=\sqrt[3]{2-\sqrt{5}}$ and the product would be: $P=(a+b)(a+bw)(a+bw^2)(aw+b)(aw+bw)(aw+bw^2)(aw^2+b)(aw^2+bw)(aw^2+bw^2)$ Where $w=\dfrac{-1+i \sqrt{3}}{2}$ is a 3rd primitive root of unity. After simplification we obtain: $P=(a^3+b^3)^3$ Finally, we simply substitute $a$ and $b$ : $P=((2+\sqrt{5})+(2-\sqrt{5}))^3=4^3=\boxed{64}$

Alan Enrique Ontiveros Salazar - 6 years, 11 months ago

I don't even understand why the real number has more than 1 value? Both radicals aren't imaginary, so it must be real and has exactly 1 value.

Samuraiwarm Tsunayoshi - 6 years, 11 months ago

The value you say is the principal value, you use it if they don't tell you anything about it. But there are other values, for example, $x=\sqrt{2}$ has a principal value of $x \approx 1.4142$ , but also has another one: $x \approx =-1.4142$ .

But here, like we have cube roots instead of square roots, each one has 3 possible values, for example: $x=\sqrt[3]{2+\sqrt{5}}$ has a principal value of $x \approx 1.618$ , and the other two this time are imaginary: $x=w\sqrt[3]{2+\sqrt{5}} \approx -0.809+1.401i$ and $x=w^2\sqrt[3]{2+\sqrt{5}} \approx -0.809-1.401i$

Alan Enrique Ontiveros Salazar - 6 years, 11 months ago

There's a difference between $x=\sqrt{2}$ and $x^2=2$ , that's what I've been taught, so I made this question.....Please tell me if I'm (w)rong.

Satvik Golechha - 6 years, 11 months ago

Yes, you are right, but here you asked us to find the product of every possible value of that expression, and since there are 2 cube roots and every cube root has 3 solutions, there are 9 possible values. Of course, the principal value, by "default" of that expression is 1, but there are another 8 values that are not principal. It can't also be 4, because there are no restriction of how to combine the roots, if you had said that that expression must satisfy the equation $x^3+3x-4=0$ , the 9 options reduces to 3, and the answer there would be 4.

Anyway, I liked this question, just readjust the wording and/or change the answer.

Alan Enrique Ontiveros Salazar - 6 years, 11 months ago

@Alan Enrique Ontiveros Salazar – @Alan Enrique Ontiveros Salazar I've changed the wordings, but I still think the correct answer would be 1, because if I ask you what is $\sqrt4$ , you should answer $2$ and not $-2$ ....

Satvik Golechha - 6 years, 11 months ago

bingo exactly how I thought

chirag shetty - 6 years, 8 months ago

elevando ao cubo essa soma chega a equação cúbica como mostrada acima. O resto é trivial e a resposta é 1.

cesar conterno - 6 years, 9 months ago

Nice one. But instead of product write find the value of . As 2 are not possible. Only 1 is. Awsome question

Nivedit Jain - 4 years, 4 months ago

An algebra problem by Satvik Golechha

The answer is 1.

1 solution

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