Question

x^{4} -26 x^{2} +25=0 equação biquadrada

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Answer 1

$x^{4}-26x^{2}+25 = 0 \\ \\ y = x^{2} \\ \\ y^{2}-26y+25=0 \\ \\ D =(-26)^{2}-4(1)(25) \\ \\ D = 676 - 100 = 576 \\ \\ \sqrt{D} = \sqrt{576} = 24 \\ \\ y' = \frac{(26+24)}{2} = 25 \\ \\ y'' = \frac{(26-24)}{2} = 1\\ \\ y = x^{2} \\ \\ 25 = x^{2} \\ \\ x = +5 \ ou \ x = -5 \\ \\ 1 = x^{2} \\ \\ x = 1 \ou \ ou x = -1$

Espero ter ajudado.

Answer 2

$\mathsf{x^4-26x^2+25=0}$

$\mathsf{a=1\quad b=-26\quad c=25}$

$\mathsf{ \Delta=b^2-4\cdot a\cdot c}$

$\mathsf{ \Delta=(-26)^2-4\cdot1\cdot25}$

$\mathsf{\Delta=6 76- 100}$

$\mathsf{ \Delta= 576}$

$\mathsf{x=\pm\sqrt{\dfrac{-b\pm\sqrt\Delta}{2\cdot a}} }$

$\mathsf{x=\pm\sqrt{\dfrac{-(-26)\pm\sqrt{576}}{2\cdot1}} }$

$\mathsf{ x=\pm\sqrt{\dfrac{26\pm24}{2}}\begin{cases}\sf x'=\sqrt{\dfrac{26+24}{2}}=\sqrt{25}=5\\\\\sf x''=-\sqrt{\dfrac{26+24}{2}}=-\sqrt{25}=-5\\\\\sf x'''=\sqrt{\dfrac{26-24}{2}}= \sqrt1=1\\\\\sf x''''=-\sqrt{\dfrac{26-24}{2}}=-\sqrt1=-1\end{cases}}$

$\boxed{\boxed{\mathsf{S=\{-5;~-1;~1;~5\}}}}$