Question

Achar as raízes das equações:
a) x² - x - 20 = 0
b) x² - 3x -4 = 0
c) x² - 8x + 7 = 0​

Answer 1

Resposta:

$x^2 - x - 20 = 0\\\\x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \left(-20\right)}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-1\right)\pm \:9}{2\cdot \:1}\\\\x_1=\frac{-\left(-1\right)+9}{2\cdot \:1}\\\\\:x_2=\frac{-\left(-1\right)-9}{2\cdot \:1}\\\\x=5\\\\\:x=-4$

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$x^2 - 3x -4 = 0\\\\x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \left(-4\right)}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-3\right)\pm \:5}{2\cdot \:1}\\\\x_1=\frac{-\left(-3\right)+5}{2\cdot \:1}\\\\\:x_2=\frac{-\left(-3\right)-5}{2\cdot \:1}\\\\x=4\\\\\:x=-1$

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$x^2-8x+7=0\\\\x_{1,\:2}=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:7}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-8\right)\pm \:6}{2\cdot \:1}\\\\x_1=\frac{-\left(-8\right)+6}{2\cdot \:1}\\\\\:x_2=\frac{-\left(-8\right)-6}{2\cdot \:1}\\\\x=7\\\\\:x=1$