Question

Equação do 2 grau- fórmula se bhaskara. A)X² + 4x - 5 = 0
B)x² + 8x + 16 = 0
C)2x² - 9x + 4 = 0
D)x² - 3x - 28 = 0

Answer 1

$a)\\x^2+4x-5=0\\\\x_{1,\:2}=\frac{-4\pm \sqrt{4^2-4\cdot \:1\cdot \left(-5\right)}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-4\pm \:6}{2\cdot \:1}\\\\x_1=\frac{-4+6}{2\cdot \:1},\:x_2=\frac{-4-6}{2\cdot \:1}\\\\x=1,\:x=-5$          $b)\\x^2+8x+16=0\\\\x_{1,\:2}=\frac{-8\pm \sqrt{8^2-4\cdot \:1\cdot \:16}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-8\pm \sqrt{0}}{2\cdot \:1}\\\\x=\frac{-8}{2\cdot \:1}\\\\x=-4\\\\x=-4$

$c)\\2x^2-9x+4=0\\\\x_{1,\:2}=\frac{-\left(-9\right)\pm \sqrt{\left(-9\right)^2-4\cdot \:2\cdot \:4}}{2\cdot \:2}\\\\x_{1,\:2}=\frac{-\left(-9\right)\pm \:7}{2\cdot \:2}\\\\x_1=\frac{-\left(-9\right)+7}{2\cdot \:2},\:x_2=\frac{-\left(-9\right)-7}{2\cdot \:2}\\\\x=4,\:x=\frac{1}{2}$               $d)\\x^2-3x-28=0\\\\x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \left(-28\right)}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-3\right)\pm \:11}{2\cdot \:1}\\\\x_1=\frac{-\left(-3\right)+11}{2\cdot \:1},\:x_2=\frac{-\left(-3\right)-11}{2\cdot \:1}\\\\x=7,\:x=-4$