Question

(x-4)(x-5) = 2x-x^2

Ayudaa​

Answer 1

Respuesta:

Raíces de x:

$x=\frac{11}{4}+i\frac{\sqrt{39}}{4},\:x=\frac{11}{4}-i\frac{\sqrt{39}}{4}$

Explicación paso a paso:

Desarrollar:

$\left(x-4\right)\left(x-5\right)=2x-x^2\\xx+x\left(-5\right)+\left(-4\right)x+\left(-4\right)\left(-5\right)=2x-x^2\\\\x^2-9x+20=2x-x^2\\\\x^2-9x+20+x^2=2x-x^2+x^2\\\\2x^2-9x+20=2x\\\\2x^2-9x+20-2x=2x-2x\\\\2x^2-11x+20=0\\\\x_{1,\:2}=\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\cdot \:2\cdot \:20}}{2\cdot \:2}\\\\*\sqrt{11^2-4\cdot \:2\cdot \:20}=\sqrt{11^2-160}=i\sqrt{160-11^2}=\sqrt{39}i\\\\x_{1,\:2}=\frac{-\left(-11\right)\pm \sqrt{39}i}{2\cdot \:2}$

$x_1=\frac{-\left(-11\right)+\sqrt{39}i}{2\cdot \:2},\:x_2=\frac{-\left(-11\right)-\sqrt{39}i}{2\cdot \:2}\\\\x=\frac{11}{4}+i\frac{\sqrt{39}}{4},\:x=\frac{11}{4}-i\frac{\sqrt{39}}{4}$