Question

How do you find the roots for `x^2 – 14x – 32 = 0`?

Answer 1

In an equation of the following form

`ax^2+bx+c=0`

the method to find the roots is:

1) calculate `Delta = b^2-4ac`
2) if `Delta=0` there is only one root `x_0=(-b)/(2a)`
3) if `Delta>0` there are two roots `x_(-)= (-b-sqrt(Delta))/(2a)`
and `x_(+) = (-b+sqrt(Delta))/(2a)`
4) if `Delta<0` there is no real solution

Example:

`x^2-14x-32=0`

`rarr a=1; b=-14; c=-32`

`rarr Delta = (-14)^2 - 4 * 1 * (-32) = 196 +128 = 324`

`Delta>0` therefore we have two roots:

`x_(-) = (14-sqrt324)/2 = (14-18)/2 = -4/2 = -2`

`x_(+) = (14+sqrt324)/2 = (14+18)/2 = 32/2 = 16`

Let us check the validity of our results:

`(-2)^2-14*(-2)-32 = 4+28-32 = 0 rarr OK`

`(16)^2-14*(16)-32 = 256-224-32 = 0 rarr OK`

Answer 2

There are several methods we could use. Here's one.

Notice that `2*16=32` and the difference between 2 and 16 is 14.

So, if the signs work out, we can factor.

`x^2-14x-32=(x+2)(x-16)`

So, `x^2-14x-32=0` if and only if

`(x+2)(x-16)=0`

Thus, we need

`x+2=0` or `x-16=0`

The solutions are:

`x=-2`, `x=16`.