Question

equacao biquadrada dois x elevada a quarta _ x elevada a segunda _15=0​

Answer 1

$Olá!!$

${\color{blue}{2x^4-x^2-15=0}}$

$Seja:\left\{\begin{array}{cc}x^4=t^2\\x^2=t\\\end{array}\right$

Logo Será:

${\color{blue}{2t^2-x-15=0}}$

$Coeficientes:\left\{\begin{array}{cc}a=2\\b=-1\\c=-15\\\end{array}\right$

Lembrando que:

$\Leftrightarrow\:\Delta=b^2-4.a.c$

$\Leftrightarrow\:\Delta=(-1)^2-4.2.(-15)$

$\Leftrightarrow\:\Delta=1+120$

$\Leftrightarrow\:\Delta=121$

$\Leftrightarrow\:x_{1},_{2}=\frac{-b±\sqrt{\Delta}}{2.a}$

$\Leftrightarrow\:t_{1},_{2}=\frac{1±\sqrt{121}}{2.2}$

$\Leftrightarrowt_{1},_{2}=\frac{1±11}{4}$

$\Leftrightarrow\:t_{1}=\frac{1+11}{4}$

$\Leftrightarrow\:t_{1}=\frac{12}{4}$

$\Leftrightarrow\:t_{1}=3$$\checkmark$

$\Leftrightarrow\:t_{2}=\frac{1-11}{4}$

$\Leftrightarrow\:t_{2}=\frac{-10}{4}$
$\Leftrightarrow\:t_{2}=-\frac{5}{2}$
NB: t2 não serve!!

$\Leftrightarrow\:x_{1},_{2}=±\sqrt{t_{1}}$

$\Leftrightarrow\:x_{1}=+\sqrt{3}$

$\Leftrightarrow\:x_{2}=-\sqrt{3}$

Answer 2

Equação biquadrada dois x elevada a quarta _ x elevada a segunda _15=0​

Explicação passo-a-passo:

2x^4 - x^2 - 15 = 0

y = x^2

2y^2 - y - 15 = 0

os coeficientes

a = 12 , b = -1 , c = -15

delta

d = 1 + 120 = 121

as raízes

y1 = (1 + 11)/4 = 3

y2 = (1 -  11)/4 = -5/2  não serve

x1 = √3

x2 = -√3