Question

X⁴-26²x²+25=0  b) x⁴-10x²=0

Answer 1

Olá,

$x^4-26x^2+25=0\\ (x^2)^2-26x^2+25=0\\\\ x^2=k\\\\ k^2-26k+25=0\\\\ \Delta=(-26)^2-4\cdot1\cdot25\\ \Delta=676-100\\ \Delta=576\\\\ k= \dfrac{-(-26)\pm \sqrt{576} }{2\cdot1} = \dfrac{26\pm24}{2}\begin{cases}k'= \dfrac{26-24}{2}= \dfrac{2}{2}=1\\\\ k''= \dfrac{26+24}{2}= \dfrac{50}{2}=25\end{cases}\\\\ k=x^2\\\\ 1=x^2\\ x=\pm \sqrt{1}\\ x=\pm1\\\\ 25=x^2\\ x=\pm \sqrt{25}\\ x=\pm5\\\\\\ \huge\boxed{\boxed{S=\{1,-1,5,-5\}}}$

_____________________

$x^4-10x^2=0\\ (x^2)^2-10x^2=0\\\\ x^2=k\\\\ k^2-10k=0\\ k(k-10)=0\\\\ k'=0~~~~k-10=0\\ ~~~~~~~~~~~~~k=10~\\\\ k=x^2\\ 0=x^2\\ x=\pm \sqrt{0}\\ x=\pm0\\\\ 10=x^2\\ x=\pm \sqrt{10}\\\\\\ \huge\boxed{\boxed{S=\{0,0, \sqrt{10},- \sqrt{10}\}}}$

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