Question

RESOLVVER LAS SIGUIENTES ECUACIONES CUADRATICAS Y DAR UNA FACTORIZACION DE CADA UNA
2. X2-7X+5=0
3. X2 + 9x + 20= 0
4. 6x + x(x – 13) = 18

Answer 1

Respuesta:

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Answer 2

Respuesta:

2. $x^2-7x+5=0$

$x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{\left(-7\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{29}}{2\cdot \:1}\\\\x_1=\frac{-\left(-7\right)+\sqrt{29}}{2\cdot \:1},\:x_2=\frac{-\left(-7\right)-\sqrt{29}}{2\cdot \:1}\\\\x=\frac{7+\sqrt{29}}{2},\:x=\frac{7-\sqrt{29}}{2}$

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3. $x^2+\:9x\:+\:20=\:0$

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

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4. $6x\:+\:x\left(x\:-\:13\right)\:=\:18$

$x^2-7x=18\\x^2-7x-18=18-18\\x^2-7x-18=0\\x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{\left(-7\right)^2-4\cdot \:1\cdot \left(-18\right)}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-7\right)\pm \:11}{2\cdot \:1}\\\\x_1=\frac{-\left(-7\right)+11}{2\cdot \:1},\:x_2=\frac{-\left(-7\right)-11}{2\cdot \:1}\\\\x=9,\:x=-2$