PILNE!!!!!!!! OBICZ a^4 + b^4 wiedząc że a^2 + b^2=9 oraz a + b=1.... Question from @isbena4

wik8947201 $\\(a^2+b^2)^2=a^4+2a^2b^2+b^4
\\.
\\(a+b)^2=a^2+2ab+b^2=1
\\2ab=1-(a^2+b^2)
\\2ab=1-9=-8/;2
\\ab=-4
\\a^4+2*(-4)^2+b^4=9^2
\\.
\\a^4+b^4=81-32
\\.
\\a^4+b^4=49
$

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animaldk $a+b=1;\ b=1-a\\\\a^2+b^2=0\\\\a^2+(1-a)^2=9\\\\a^2+1-2a+a^2-9=0\\\\2a^2-2a-8=0\ /:2\\\\a^2-a-4=0\\\\\Delta=(-1)^2-4\cdot1\cdot(-4)=1+16=17;\ \sqrt\Delta=\sqrt{17}$

$a_1=\frac{1-\sqrt{17}}{2};\ a_2=\frac{1+\sqrt{17}}{2}\\\\b_1=1-\frac{1-\sqrt{17}}{2}=\frac{1+\sqrt{17}}{2};\ b_2=1-\frac{1+\sqrt{17}}{2}=\frac{1-\sqrt{17}}{2}\\\\a_1=b_2\ \wedge\ a_2=b_1$

$a^4+b^4=(a^2)^2+(b^2)^2\\\\a^2=(\frac{1-\sqrt{17}}{2})^2=\frac{1-2\sqrt{17}+17}{4}=\frac{18-2\sqrt{17}}{4}=\frac{9-\sqrt{17}}{2}\\\\a^4=(\frac{9-\sqrt{17}}{2})^2=\frac{81-18\sqrt{17}+17}{4}=\frac{98-18\sqrt{17}}{4}=\frac{49-9\sqrt{17}}{2}\\\\b^2=(\frac{1+\sqrt{17}}{2})^2=\frac{1+2\sqrt{17}+17}{4}=\frac{18+2\sqrt{17}}{4}=\frac{9+\sqrt{17}}{2}\\\\b^4=(\frac{9+\sqrt{17}}{2})^2=\frac{81+18\sqrt{17}+17}{4}=\frac{98+18\sqrt{17}}{4}=\frac{49+9\sqrt{17}}{2}$

$a^4+b^4=\frac{49-9\sqrt{17}}{2}+\frac{49+9\sqrt{17}}{2}=\frac{98}{2}=49$

©DRK

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