Eigenvalue wrote:x²=16, then x=[tex]\pm[/tex]4
There are 2 cases, x=4 and x=-4
If x=4
(4)³(y)²=216
y²=[tex]\frac{27}{8}[/tex]
y=[tex]\frac{3 \sqrt{3} }{2 \sqrt{2} }[/tex]
y=[tex]\frac{3 \sqrt{6} }{4}[/tex]
x-y=4-[tex]\frac{3 \sqrt{6} }{4}[/tex]
If x=-4
(-4)³(y)²=216
y²=[tex]\frac{-27}{8}[/tex]
y=[tex]\frac{-3 \sqrt{3} }{2 \sqrt{2} }[/tex]
y=[tex]\frac{-3 \sqrt{6} }{4}[/tex]
x-y=4+[tex]\frac{3 \sqrt{6} }{4}[/tex]
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