Question

How do you solve `x^2 + 2x - 15 = 0`?

Answer 1

`x=-5" or "x=3`

Explanation:

`"the factors of - 15 which sum to + 2 are + 5 and - 3"`

`rArr(x+5)(x-3)=0`

`"equate each factor to zero and solve for x"`

`x+5=0rArrx=-5`

`x-3=0rArrx=3`

Answer 2

`x=-5 or 3`

Explanation:

Rules for factorising:

• `ax^2+bx+c=0`
• `m xx n = c`
• `m + n = b`
• `:.(x+m)(x+n) = 0`

In this case:

`5 xx -3 = -15 = c`

`5 + -3 = 2 = b`

`:. x^2+2x−15=0 -> color(red)((x+5)(x-3) = 0)`

Either of the brackets must be equal to 0.

Assuming (x+5) = 0:

`color(red)(x=-5)`

Assuming (x-3) = 0:

`color(red)(x=3)`