Apr 17, 2013 Thread starter Admin #1 anemone MHB POTW Director Staff member Feb 14, 2012 3,815 For what integers p and q is where \(\displaystyle x=\sqrt {29}+\sqrt {89}\) is a root of the equation \(\displaystyle x^4+px^2+q=0\)

For what integers p and q is where \(\displaystyle x=\sqrt {29}+\sqrt {89}\) is a root of the equation \(\displaystyle x^4+px^2+q=0\)

Apr 18, 2013 Moderator #2 Opalg MHB Oldtimer Staff member Feb 7, 2012 2,740 anemone said: For what integers p and q is where \(\displaystyle x=\sqrt {29}+\sqrt {89}\) is a root of the equation \(\displaystyle x^4+px^2+q=0\) Click to expand... Spoiler 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 $p=-236,\ q=3600.$

anemone said: For what integers p and q is where \(\displaystyle x=\sqrt {29}+\sqrt {89}\) is a root of the equation \(\displaystyle x^4+px^2+q=0\) Click to expand... Spoiler 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 $p=-236,\ q=3600.$