Question

equação de 2° grau -x²+8x+9=0

Answer 1

-x²+8x+9=0

a = -1
b = 8
c = 9

Δ = b² -4 .a .c
Δ = 8² -4 .(-1) . 9
Δ = 64 + 36
Δ = 100

-b +- √Δ    /2.a
-8 +- √100    /2.-1
-8 +- 10     / -2

x' =  -8 + 10   / -2
x' = 2  / -2
x' = -1

x" = -8 - 10   /-2
x" = -18 / -2
x" = 9

Resposta: S:{ -1  e 9 }

Answer 2

Olá Jaisarosa.


Resolvendo por Bhaskara:

$\mathsf{-x^2+8x+9=0}\\\\\\\\\mathsf{\Delta=8^2-4\cdot(-1)\cdot9}\\\mathsf{\Delta=64+36}\\\mathsf{\Delta=100}\\\\\\\\\mathsf{x=\dfrac{-8\pm\sqrt{100}}{2\cdot(-1)}}\\\\\\\mathsf{x^+=\dfrac{-8+10}{-2}\qquad\qquad\qquad\qquad x^-=\dfrac{-8-10}{-2}}\\\\\\\mathsf{x^+=\dfrac{2}{-2}\qquad\qquad\qquad\qquad\qquad~ x^-=\dfrac{-18}{-2}}\\\\\\\boxed{\mathsf{x^+=-1}}\qquad\qquad\qquad\qquad~\quad\boxed{\mathsf{x^-=9}}$


Resolvendo pelo método de completar quadrados:


$\mathsf{-x^2+8x+9=0~\cdot(-1)}\\\\\mathsf{x^2-8x-9=0+(4^2)}\\\\\mathsf{x^2-8x+16=16+9}\\\\\mathsf{(x-4)^2=25}\\\\\mathsf{\sqrt{(x-4)^2}=\pm \sqrt{25}}\\\\\mathsf{x-4=\pm5}\\\\\mathsf{x=5+4}\\\\\boxed{\mathsf{x=9}}\\\\\\\mathsf{x=-5+4}\\\\\boxed{\mathsf{x=-1}}$


Dúvidas? comente.