Question

O produto das raízes positivas de x^4-11x^2+18=0?

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo-a-passo:

$\sf x^4 - 11x^2 + 18 = 0$

$\sf x^2 = y$

$\sf y^2 - 11y + 18 = 0$

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

$\sf \Delta = (-11)^2 - 4.(1).(18)$

$\sf \Delta = 121 - 72$

$\sf \Delta = 49$

$\sf y = \dfrac{-b \pm \sqrt{\Delta}}{2a}$

$\sf y' = \dfrac{11 + \sqrt{49}}{2.1} = \dfrac{11 + 7}{2} = \dfrac{18}{2} = 9$

$\sf y'' = \dfrac{11 - \sqrt{49}}{2.1} = \dfrac{11 - 7}{2} = \dfrac{4}{2} = 2$

$\sf (x')^2 = 9$

$\sf x' = 3$

$\sf (x'')^2 = 2$

$\sf x'' = \sqrt{2}$

$\sf P = x'.x''$

$\sf P = 3.\sqrt{2}$

$\boxed{\boxed{\sf P = 3\sqrt{2}}}$

Answer 2

$\sf x^4-11x^2+18=0$

$\sf (x^2)^2-11x^2+18=0$

Seja $\sf k=x^2$

$\sf k^2-11k+18=0$

$\sf \Delta=(-11)^2-4\cdot1\cdot18$

$\sf \Delta=121-72$

$\sf \Delta=49$

$\sf k=\dfrac{-(-11)\pm\sqrt{49}}{2\cdot1}=\dfrac{11\pm7}{2}$

$\sf k'=\dfrac{11+7}{2}~\Rightarrow~k'=\dfrac{18}{2}~\Rightarrow~k'=9$

$\sf k"=\dfrac{11-7}{2}~\Rightarrow~k"=\dfrac{4}{2}~\Rightarrow~k"=2$

Para $\sf k=9$:

$\sf x^2=9$

$\sf x=\pm\sqrt{9}$

• $\sf x'=3$

• $\sf x"=-3$

Para $\sf k=2$:

$\sf x^2=2$

$\sf x=\pm\sqrt{2}$

• $\sf x'=\sqrt{2}$

• $\sf x"=-\sqrt{2}$

O produto das raízes positivas é $\sf \red{3\sqrt{2}}$