Question

PLEASE HURRY
use the intercept method to solve the equation x-2=x^2-4

Answer 1

`\Large \boxed{x=2} \\ \\ \Large \boxed{x=-1}`

Explanation:

`x-2=x^2-4`

Subtract (x² - 4) from both sides.

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

Distribute negative sign.

`x-2-x^2+4=x^2-4-x^2+4`

Combine like terms.

`-x^2 +x+2=0`

Factor left side of the equation.

`-x^2 -x+2x+2=0`

`-x(x+1)+2(x+1)`

Take (x+1) as a common factor.

`(-x+2)(x+1)`

Set factors equal to 0.

`-x+2=0 \\ \\ -x=-2 \\ \\ x=2`

`x+1 \\ \\ x=-1`

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

Answer 2

`x_1=2`

`x_2=-1`

Explanation:

`x-2=x^2-4`

`0=x^2-x-4+2`

`0=x^2-x-2`

Let's find x-intercepts. It is when
`y=0`

I will use the quadratic equation

`$x=(-b\pm √(b^2-4ac))/(2a)$`

`$x=(-\left(-1\right)\pm√(\left(-1\right)^2-4\cdot \:1\cdot \left(-2\right)))/(2\cdot \:1)$`

`$x=(1\pm√(9))/(2)=(1 \pm 3)/(2) $`

`x_1=2`

`x_2=-1`