Question

What are the x-intercepts of `x²-y+2x=6`?

Answer 1

The x-intercepts are at `( -1+sqrt7,0)` and `( -1-sqrt7,0)`

Explanation:

We first isolate y.

`x^2-y+2x=6`
`-y+2x=6-x^2`
`-y=6-x^2-2x`
`y=-6+x^2+2x`
`y=x^2+2x-6`

We now try to find the zeros. Since we cannot factor this equation in any quick way, we find the two zeros using the quadratic formula `(-b+-sqrt(b^2-4(a)(c)))/(2(a))`
`a=1`
`b=2`
`c=-6`

The zeros are:
`(-2+-sqrt(2^2-4(1)(-6)))/(2(1))`
`(-2+-sqrt(4+24))/2`
`(-2+-sqrt28)/2`
`(-2+-2sqrt7)/2`
`-1+-sqrt7`

`-1+-sqrt7` are the x values for the x-intercepts.

Note that we know that(and it will, if you are correct) when we plug in the zeros, we get zero for the equation.

*Here is the idea of how the zeros are the answer. Since the y value is 0, we know that the point must lie on the x coordinate, which means those points are the x intercepts.