How do you solve #x^2-x=2#?

2 Answers
Jun 28, 2015

Answer:

Solve #y = x^2 - x - 2 = 0#

Explanation:

Since (a - b + c = 0), use shortcut -> One real root is (-1) and the other is (-c/a = 2)

Jun 29, 2015

Answer:

The solutions are #2# and #-1#.

Explanation:

To solve: #x^2 -x =2#,

First we note that this is a quadratic equation. We are used to seeing quadratic equations as: #ax^2 + bx + c = 0#, so let's make the equation in this question look like that:

#x^2 -x =2#, Subtract #2# (add #-2#) on both sides)

#x^2 -x -2 = 0#.

Now we have choices, we could try to factor, we could complete the square, we could use the quadratic formula. I like to try to factor first, because if the quadratic factors easily, I find that fastest. But don't spend a lot of time factoring, because we have 2 other methods that will work even if we can't easily find the factorization.

It is fairly straightforward to factor: #x^2 -x -2#. We get:

#(x-2) (x +1) = 0#.

That leads to

#x-2 =0# #color(white)"xx"# or #color(white)"xx"# #x+1 =0#

#x = 2# #color(white)"xx"# or #color(white)"xx"# #x = -1#