Question

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

Answer 1

x⁴ - 26x² + 25 = 0

x⁴ = (x²)² = y²
x² = y

y² - 26y + 25 = 0

a = 1; b = -26; c = 25

y = [- b ± √(b² - 4ac)] / 2a
y = [- (-26) ± √([-26]² - 4 . 1 . 25)] / 2 . 1
y = [26 ± √(676 - 100)] / 2
y = [26 ± √576] / 2
y = [26 ± 24] / 2
y' = [26 - 24] / 2 = 2 / 2 = 1
y'' = [26 + 24] / 2 = 50 / 2 = 25

Como x² = y, temos:
x² = 1                   x² = 25
x = ± √1               x = ± √25
x = ± 1                  x = ± 5

S = {-5, -1, 1, 5}

Espero ter ajudado. Valeu!

Answer 2

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

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

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

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

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

$\mathsf{ \Delta= 576}$

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

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

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

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