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Here is what I did

$$x + \sqrt {17 - x^2} + x\times\sqrt{17 - x^2} = 9$$ I can`t undestand how to solve it, any help would be appreciated!

riveges755
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    Welcome to MSE, Please add where are you stuck and what you know so we can help you with the problem and for the question doesn't get closed. – mark Dec 20 '21 at 13:43
  • Formatting note: if you want the square root to encompass a long expression, enclose the expression in curly brackets. Thus \sqrt {17-x^2} renders as $\sqrt {17-x^2}$. – lulu Dec 20 '21 at 13:57
  • Thank you for useful tips! – riveges755 Dec 20 '21 at 13:59
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    Note: in this case there are small integer solutions, so perhaps you were meant to solve this numerically from the start (or at least to apply the rational root theorem to the quartic polynomial one gets by squaring). – lulu Dec 20 '21 at 14:02
  • I need a symmetric polynomials way, my teacher demands, but doesn`t explain how to do it. – riveges755 Dec 20 '21 at 14:09
  • Thank you for adding your attempt. For future questions, please use MathJax. Click on "Edit" to see what lulu did to make your equation render properly. – user21820 Dec 21 '21 at 08:38

1 Answers1

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By using your substitution ($y=\sqrt{17-x^2}$), we get

$$ \begin{cases} {x+y+xy=9} \\ {x^2+y^2=17} \end{cases} $$

From the first equation we have,

$$(x+y)^2=(9-xy)^2\;\Rightarrow\ 17+2xy=(xy)^2-18(xy)+81$$ And by solving the quadratic equation in $xy$ we have $xy\in\{16,4\}$. Hence there are two cases, $$ \begin{cases} {xy=16} \\ {x+y=-7} \end{cases} \qquad \text{or} \qquad \begin{cases} {xy=4} \\ {x+y=5} \end{cases} $$

Can you finish it now?

Etemon
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