# How do you solve #x^2+2x-48=0#?

##### 2 Answers

We can factor the quadratic to obtain

#### Explanation:

One way to solve this equation without relying on the Quadratic Formula is to factor the quadratic. Once we have the factors, we know that setting either of them equal to zero will make the whole expression zero thus satisfying the equation.

Let's imagine we know our factors, and let's call them

Expanding the factored portion we get

First, looking at the

To factor this quadratic we need to know if there are factors that satisfy the following - they multiply to give

so the numbers we are looking for must be next door to

Setting each factor equal to zero we can solve for

So the solutions are

A quick graph of the function confirms our calculations:

graph{x^2+2x-48 [-10, 10, -60, 40]}

6 and -8

#### Explanation:

We can directly find the 2 real roots without having to factor the

trinomial. Use the new Transforming Method (Socratic Search)

y = x^2 + 2x - 48.

Find 2 numbers (real roots) knowing sum (-b = -2) and product (c = -48). The 2 real roots have opposite sign because ac < 0.

Factor pairs of (c = -48) --> (-4, 12)(4 - 12)(-6, 8)(6, -8). This last sum is (-2 = -b). There for, the 2 real roots are: 6 and -8.