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

1 Answer
Parabola
Dec 5, 2017

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.

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