Respuesta :

mirai123

Answer:

[tex]x_1=2[/tex]

[tex]x_2=-1[/tex]

Step-by-step explanation:

[tex]x-2=x^2-4[/tex]

[tex]0=x^2-x-4+2[/tex]

[tex]0=x^2-x-2[/tex]

Let's find x-intercepts. It is when [tex]y=0[/tex]

I will use the quadratic equation

[tex]$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$[/tex]

[tex]$x=\frac{-\left(-1\right)\pm\sqrt{\left(-1\right)^2-4\cdot \:1\cdot \left(-2\right)}}{2\cdot \:1}$[/tex]

[tex]$x=\frac{1\pm\sqrt{9}}{2}=\frac{1 \pm 3}{2} $[/tex]

[tex]x_1=2[/tex]

[tex]x_2=-1[/tex]

09pqr4sT

Answer:

[tex]\Large \boxed{x=2} \\ \\ \Large \boxed{x=-1}[/tex]

Step-by-step explanation:

[tex]x-2=x^2-4[/tex]

Subtract (x² - 4) from both sides.

[tex]x-2-(x^2-4)=x^2-4-(x^2-4)[/tex]

Distribute negative sign.

[tex]x-2-x^2+4=x^2-4-x^2+4[/tex]

Combine like terms.

[tex]-x^2 +x+2=0[/tex]

Factor left side of the equation.

[tex]-x^2 -x+2x+2=0[/tex]

[tex]-x(x+1)+2(x+1)[/tex]

Take (x+1) as a common factor.

[tex](-x+2)(x+1)[/tex]

Set factors equal to 0.

[tex]-x+2=0 \\ \\ -x=-2 \\ \\ x=2[/tex]

[tex]x+1 \\ \\ x=-1[/tex]

The solutions to the equation are x = 2 or x = -1.

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