Question

Como se resolve essa equação (5x-1)²+12=21

Answer 1

25x²-10x+1+12=21
25x²-10x-8=0 (Bhaskara)

$\frac{10+- \sqrt{(10)^2-4*25*(-8)} }{2*(25)} = \frac{10+- \sqrt{900} }{50} = \frac{10+30}{50}$

x1= 4/5
x2= -2/5

Answer 2

Olá.

Primeiro vamos aplicar produtos notáveis em (5x-1)² e logo após resolveremos a equação de segundo grau.

$\mathsf{(5x-1)^2+12=21}\\\\ \mathsf{(5^2x^2-1\cdot5x\cdot2+(-1)^2)+12=21}\\\\ \mathsf{(25x^2-10x+1)+12=21}\\\\ \mathsf{25x^2-10x+1+12=21}\\\\ \mathsf{25x^2-10x+13-21=0}\\\\ \boxed{\mathsf{25x^2-10x-8=0}}$

Vamos agora resolver a equação de 2° grau.
$\left\{\begin{array}{c}\mathsf{a=25}\\\mathsf{b=-10}\\\mathsf{c=-8}\\ \end{array}\right\\\\\\ \mathsf{x=\dfrac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2a}}\\\\\\ \mathsf{x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\cdot25\cdot(-8)}}{2\cdot25}}\\\\\\ \mathsf{x=\dfrac{10\pm\sqrt{100-4\cdot(-200)}}{50}}\\\\\\ \mathsf{x=\dfrac{10\pm\sqrt{100+800}}{50}}\\\\\\ \mathsf{x=\dfrac{10\pm\sqrt{900}}{50}}\\\\\\ \mathsf{x=\dfrac{10\pm30}{50}}$

Vamos descobrir agora as duas raízes:
$\mathsf{x=\dfrac{10+30}{50}}\\\\\\ \mathsf{x=\dfrac{40}{50}}\\\\\\ \mathsf{x=\left(\dfrac{40}{50}\right)^{:10}}\\\\\\ \boxed{\mathsf{x=\dfrac{4}{5}=0,8}}\\\\\\\\\\ \mathsf{x=\dfrac{10-30}{50}}\\\\\\ \mathsf{x=\dfrac{-20}{50}}\\\\\\ \mathsf{x=\left(\dfrac{-20}{50}\right)^{:10}}\\\\\\ \boxed{\mathsf{x=\dfrac{-2}{5}=-0,4}}$

Qualquer dúvida, deixe nos comentários.
Bons estudos.