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Here is the problem:

Three numbers have a sum of $5$ and the sum of their squares is $29$. If the product of the three numbers is $−10$, what are the three numbers? Express your answer in simplest radical form.

I used Vieta's formulas to get $x^3-5x^2+12x+10=0$

What should I do after this?

KingLogic
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  • No, I'm not allowed and also when I graph the equation, I only get one value for x –  Sep 21 '20 at 23:25
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    Your calculation of the coefficient before $x$ is wrong. It should be $\frac 1 2(5^2- 29) = -2$. Therefore the equation reads $x^3 - 5x^2 - 2x + 10 = 0$, and the three roots are $5, \pm\sqrt 2$. – WhatsUp Sep 21 '20 at 23:26

1 Answers1

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As @WhatsUp noted in the comments, the equation should have been $x^3-5x^2-2x+10=0$, and so you can find by inspection or by the Rational Root Theorem that $x=5$ is a root. Divide and get $x^2-2=0$, and so your answers should be $x=5, \sqrt{2}, -\sqrt{2}$.

Edit: Another (essentially equivalent, but more detailed) method to get the coefficient of $-2$ for the linear term:

Assume the roots are $a, b, c$.

Square the first equation: $(a+b+c=5)^2\Rightarrow (a^2+b^2+c^2+2(ab+ac+bc))=25$

We know that $a^2+b^2+c^2=29$ by the second equation, and so we get $2(ab+ac+bc)=-4\Rightarrow ab+ac+bc=-2$

KingLogic
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