Question

resolva fórmula de bhaskara
a) x2 -4x-5=0

b) -x2 +x+12=0
​

Answer 1

Sua questão:

resolva fórmula de bhaskara

a) x2 -4x-5=0

b) -x2 +x+12=0

Resolução:

$\large\begin{array}{lr} \sf x^2 -4x-5=0\left\{\begin{array}{ll}\sf a=1\\\sf b=-4\\\sf c=-5\end{array}\right. \end{array}$

$\large\begin{array}{lr} \sf \Delta =b^2 -4 *a*c\\\\\sf \Delta = (-4)^2 -4*1*(-5)\\\\\sf \Delta = 16 + 20\\\\\sf \underline{\boxed{\red{\sf \Delta =36}}} \end{array}$

$\large\begin{array}{lr} \sf \sf x= \dfrac{-b \pm \sqrt{\Delta} }{2*a}\\\\\sf x= \dfrac{-(-4) \pm \sqrt{36} }{2*1} \\\\ \sf x= \dfrac{ 4 \pm 6 }{2} \\\\ \sf x'= \dfrac{10}{2} = \underline{\boxed{\red{\sf x' = 5}}} \\\\ \sf x''= \dfrac{-2}{2} = \underline{\boxed{\red{\sf x'' = -1}}} \end{array}$

__________#___________

$\large\begin{array}{lr} \sf -x^2 +x+12=0\left\{\begin{array}{ll}\sf a=-1\\\sf b=1\\\sf c=12\end{array}\right. \end{array}$

$\large\begin{array}{lr} \sf \Delta =b^2 -4 *a*c\\\\\sf \Delta = 1^2 -4*(-1)*12\\\\\sf \Delta = 1 + 48\\\\\sf \underline{\boxed{\red{\sf \Delta =49}}} \end{array}$

$\large\begin{array}{lr} \sf \sf x= \dfrac{-b \pm \sqrt{\Delta} }{2*a}\\\\\sf x= \dfrac{-1 \pm \sqrt{49} }{2*(-1)} \\\\ \sf x= \dfrac{ -1 \pm 7 }{-2} \\\\ \sf x'= \dfrac{6}{-2} = \underline{\boxed{\red{\sf x' = -3}}} \\\\ \sf x''= \dfrac{-8}{-2} = \underline{\boxed{\red{\sf x'' = 4}}} \end{array}$

Espero ter ajudado.

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