Question

How do you factor and solve `x^2+2x-8=0`?

Answer 1

The solutions for the equation are `color(blue)(x=2 , color(blue)(x=-4`

Explanation:

`x^2 + 2x - 8 = 0`

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like `ax^2 + bx + c`, we need to think of 2 numbers such that:

`N_1*N_2 = a*c = 1*(-8) = -8`

AND

`N_1 +N_2 = b = 2`

After trying out a few numbers we get `N_1 = -2` and `N_2 =4`
`(-2)*4 = -8`, and `-2+ 4= 2`

`x^2 + color(blue)( 2x) - 8 = x^2 + color(blue)(4x-2x) - 8`

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

`color(blue)((x+4))` is a common factor to each of the terms

`color(green)(( x - 2 )(x+4)` is the factorised form of the expression. Now we equate the factors to zero.

• `color(blue)(x-2) =0 , color(blue)(x=2`

• `color(blue)(x + 4) =0 , color(blue)(x=-4`

Answer 2

2 and -4

Explanation:

Solve `y = x^2 + 2x - 8.`
The 2 real roots have opposite signs because ac < 0
Find 2 real roots knowing sum (-b = -2) and product (c = -8).
Factor pairs of (c = -8) --> (-2, 4)(2, -4). This last sum is (-2 = -b).
Therefor, the 2 real roots are: 2 and -4.