Question

Resolva as equações quadráticas usando a fórmula de Bhaskara.
a) - 2 - 6x + 16 = 0
b) - x2 + x + 2 = 0
c) x2 - 12x + 35 = 0​

Answer 1

Resposta:

a) x₁ = -8, x₂ = 2

b) x₁ = -1, x₂ = 2

c) x₁ = 7, x₂ = 5

Explicação passo-a-passo:

a) $-x^2-6x+16=0$

$x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\left(-1\right)\cdot \:16}}{2\left(-1\right)}\\x_{1,\:2}=\frac{-\left(-6\right)\pm \:10}{2\left(-1\right)}\\x_1=\frac{-\left(-6\right)+10}{2\left(-1\right)},\:x_2=\frac{-\left(-6\right)-10}{2\left(-1\right)}$

$x_1=\frac{-\left(-6\right)+10}{2\left(-1\right)}=\frac{6+10}{-2\cdot \:1}=\frac{16}{-2\cdot \:1}=\frac{16}{-2}=-\frac{16}{2}=-8$

$x_2=\frac{-\left(-6\right)-10}{2\left(-1\right)}=\frac{6-10}{-2\cdot \:1}=\frac{-4}{-2\cdot \:1}=\frac{-4}{-2}=\frac{4}{2}=2$

$x_1=-8,\:x_2=2$

b) $-x^2+x+2=0$

$x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\left(-1\right)\cdot \:2}}{2\left(-1\right)}\\x_{1,\:2}=\frac{-1\pm \:3}{2\left(-1\right)}\\x_1=\frac{-1+3}{2\left(-1\right)},\:x_2=\frac{-1-3}{2\left(-1\right)}$

$x_1=\frac{-1+3}{2\left(-1\right)}=\frac{-1+3}{-2\cdot \:1}=\frac{2}{-2\cdot \:1}=\frac{2}{-2}=-\frac{2}{2}=-1$

$x_2=\frac{-1-3}{2\left(-1\right)}=\frac{-1-3}{-2\cdot \:1}=\frac{-4}{-2\cdot \:1}=\frac{-4}{-2}=\frac{4}{2}=2$

$x=-1,\:x=2$

c) $x^2-12x+35=0$

$x_{1,\:2}=\frac{-\left(-12\right)\pm \sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:35}}{2\cdot \:1}\\x_{1,\:2}=\frac{-\left(-12\right)\pm \:2}{2\cdot \:1}\\x_1=\frac{-\left(-12\right)+2}{2\cdot \:1},\:x_2=\frac{-\left(-12\right)-2}{2\cdot \:1}$

$x_1=\frac{-\left(-12\right)+2}{2\cdot \:1}=\frac{12+2}{2\cdot \:1}=\frac{14}{2\cdot =7\:1}=\frac{14}{2}$

$x_2=\frac{-\left(-12\right)-2}{2\cdot \:1}=\frac{12-2}{2\cdot \:1}=\frac{10}{2\cdot \:1}=\frac{10}{2}=5$

$x_1=7,\:x_2=5$