Find the integer values of p and q.

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SUMMARY

The integers p and q for which \( x = \sqrt{29} + \sqrt{89} \) is a root of the equation \( x^4 + px^2 + q = 0 \) are determined through substitution and simplification. By calculating \( x^2 \), we find \( x^2 = 29 + 89 + 2\sqrt{29 \cdot 89} = 118 + 2\sqrt{2581} \). Subsequently, substituting \( x^2 \) into the polynomial reveals that \( p = -118 \) and \( q = -2581 \). Thus, the solution is \( p = -118 \) and \( q = -2581 \).

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For what integers p and q is where $$x=\sqrt {29}+\sqrt {89}$$ is a root of the equation $$x^4+px^2+q=0$$
 
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anemone said:
For what integers p and q is where $$x=\sqrt {29}+\sqrt {89}$$ is a root of the equation $$x^4+px^2+q=0$$
If $x=\sqrt {29}+\sqrt {89}$ then $x^2 = 29+89 + 2\sqrt{29*89} = 118 + 2\sqrt{29*89}$, and $(x^2 - 118)^2 = 4*29*89$. That is, $x^4 - 236x^2 + (118^2 - 4*29*89) = 0$, or $x^4 - 236 x^2 + 3600 = 0.$ So $p=-236,\ q=3600.$
 

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