Question

1. Resolva as equações do 2º grau em R. a) x² + 10x + 25=0 b) 3x² + 4x + 1=0 c) x² + 12x - 20=0 d) x² + 5x +6=0 e) x² - 7x + 12=0​

Answer 1

Resposta:

a) x = -5

b) x' = -2 / 6; x'' = -1

c) $x=2\left(\sqrt{14}-3\right),\:x=-2\left(3+\sqrt{14}\right)$

d) x' = -2 ; x'' = -3

e) x' = 4 ; x'' = 3

Explicação passo-a-passo:

a) Calculando o Δ da equação completa:

Δ = b2 - 4.a.c

Δ = 102 - 4 . 1 . 25

Δ = 100 - 4. 1 . 25

Δ = 0

Há 1 raiz real.

Neste caso, x' = x'':

Aplicando Bhaskara:

x = (-b +- √Δ)/2a

x' = (-10 + √0)/2.1    

x' = -10 / 2    

x' = -5    

x'' = (-10 - √0)/2.1

x'' = -10 / 2

x'' = -5

b) Calculando o Δ da equação completa:

Δ = b2 - 4.a.c

Δ = 42 - 4 . 3 . 1

Δ = 16 - 4. 3 . 1

Δ = 4

Há 2 raízes reais.

x' = (-4 + √4)/2.3    

x' = -2 / 6    

x'' = (-4 - √4)/2.3

x'' = -6 / 6

x'' = -1

c) $\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$    

$\mathrm{Para\:}\quad a=1,\:b=12,\:c=-20:\quad x_{1,\:2}=\frac{-12\pm \sqrt{12^2-4\cdot \:1\left(-20\right)}}{2\cdot \:1}$

$\frac{-12+\sqrt{12^2-4\cdot \:1\cdot \left(-20\right)}}{2\cdot \:1}$ : $2\left(\sqrt{14}-3\right)$  e $\frac{-12-\sqrt{12^2-4\cdot \:1\cdot \left(-20\right)}}{2\cdot \:1}$ : $-2\left(3+\sqrt{14}\right)$

$x=2\left(\sqrt{14}-3\right),\:x=-2\left(3+\sqrt{14}\right)$

d) $a=1,\:b=5,\:c=6:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:1\cdot \:6}}{2\cdot \:1}$

$\frac{-5+\sqrt{5^2-4\cdot \:1\cdot \:6}}{2\cdot \:1}$ = -2

$\frac{-5-\sqrt{5^2-4\cdot \:1\cdot \:6}}{2\cdot \:1}$ = -3

e) $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{Para\:}\quad a=1,\:b=-7,\:c=12:\quad x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{\left(-7\right)^2-4\cdot \:1\cdot \:12}}{2\cdot \:1}$

$\frac{-\left(-7\right)+\sqrt{\left(-7\right)^2-4\cdot \:1\cdot \:12}}{2\cdot \:1}$ = 4

$\frac{-\left(-7\right)-\sqrt{\left(-7\right)^2-4\cdot \:1\cdot \:12}}{2\cdot \:1}$ = 3