Question

1) resolva em R

A) x²+4x-96=0

B) x²+11x+40=0

C) x²-14x49=0

Answer 1

Vamos là.

A)

x² + 4x - 96 = 0

delta

d = 16 + 4*96 = 400

x1 = (-4 + 20)/2 = 16/2 = 8

x2 = (-4 - 20)/2 = -24/2 = -12

B)

x² + 11x + 40 = 0

delta negativo não ha raiz real.

C)

x² - 14x + 49 = 0

(x - 7)² = 0

x = 7

Answer 2

Explicação passo a passo:

$x^2+4x-96=0\\\\x_{1,\:2}=\frac{-4\pm \sqrt{4^2-4\cdot \:1\cdot \left(-96\right)}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-4\pm \:20}{2\cdot \:1}\\\\x_1=\frac{-4+20}{2\cdot \:1},\:x_2=\frac{-4-20}{2\cdot \:1}\\\\S[8,-12]$

$x^{2} +11x+40=0\\\\x_{1,\:2}=\frac{-11\pm \sqrt{11^2-4\cdot \:1\cdot \:40}}{2\cdot \:1}\\\\_{1,\:2}=\frac{-11\pm \sqrt{39}i}{2\cdot \:1}\\x_1=\frac{-11+\sqrt{39}i}{2\cdot \:1},\:x_2=\frac{-11-\sqrt{39}i}{2\cdot \:1}\\\\\\x=-\frac{11}{2}+i\frac{\sqrt{39}}{2},\:x=-\frac{11}{2}-i\frac{\sqrt{39}}{2}$

$x^2-14x+49=0\\\\x_{1,\:2}=\frac{-\left(-14\right)\pm \sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:49}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-14\right)\pm \sqrt{0}}{2\cdot \:1}\\\\x=\frac{-\left(-14\right)}{2\cdot \:1}\\\\x=7$