Question

x²-8x=0 y²+10y+24=0. (x-4)(x-3)=8 ayúdenme es que yo no le entiendo a esto y nesecito mucha ayuda seño suplico enserio ​

Answer 1

Respuesta:

:-)

Explicación paso a paso:

a)

$x^2-8x=0\\\\Resolver\ con\ la\ formula\ general\ para\ ecuaciones\ de\ segundo\ grado\ :\\\\x_{1,\:2}=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:0}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-8\right)\pm \:8}{2\cdot \:1}\\\\x_1=\frac{-\left(-8\right)+8}{2\cdot \:1},\:x_2=\frac{-\left(-8\right)-8}{2\cdot \:1}\\\\\\x=8,\:x=0$

b)

$y^2+10y+24=0\\\\Resolver\ con\ la\ formula\ general\ para\ ecuaciones\ de\ segundo\ grado\ :\\\\y_{1,\:2}=\frac{-10\pm \sqrt{10^2-4\cdot \:1\cdot \:24}}{2\cdot \:1}\\\\y_{1,\:2}=\frac{-10\pm \:2}{2\cdot \:1}\\\\y=\frac{-10+2}{2\cdot \:1},\:y_2=\frac{-10-2}{2\cdot \:1}\\\\\\\y=-4,\:y=-6$

c)

$\left(x-4\right)\left(x-3\right)=8\\\\x^2-7x+12=8\\\\x^2-7x+12-8=0\\\\Resolver\ con\ la\ formula\ general\ para\ ecuaciones\ de\ segundo\ grado\ :\\\\x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{\left(-7\right)^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{33}}{2\cdot \:1}\\\\x_1=\frac{-\left(-7\right)+\sqrt{33}}{2\cdot \:1},\:x_2=\frac{-\left(-7\right)-\sqrt{33}}{2\cdot \:1}\\\\\\x=\frac{7+\sqrt{33}}{2},\:x=\frac{7-\sqrt{33}}{2}$