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

May 11, 2016

We can factor the quadratic to obtain $x = - 8 , + 6$

#### 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 $a$ and $b$ and a common factor $c$. We will end up with an equation in the form

$c \left(x + a\right) \left(x + b\right) = {x}^{2} + 2 x - 48 = 0$

Expanding the factored portion we get

$c {x}^{2} + c a x + c b x + c a b = {x}^{2} + 2 x - 48 = 0$

First, looking at the ${x}^{2}$ terms, we see that our common factor, $c = 1$. Now we can pull the terms in $x$ together and re-write our expression as

${x}^{2} + \textcolor{m a \ge n t a}{\left(a + b\right)} x + \textcolor{b l u e}{a b} = {x}^{2} + \textcolor{m a \ge n t a}{2} x \textcolor{b l u e}{- 48} = 0$

To factor this quadratic we need to know if there are factors that satisfy the following - they multiply to give $\textcolor{b l u e}{- 48}$ and add to give $\textcolor{m a \ge n t a}{+ 2}$. This implies that the factors of $48$ that we are looking for are integers which are separated by $2$. This must be on either side of the square-root of our number.

$\sqrt{48} \cong 6.928$

so the numbers we are looking for must be next door to $7$, which implies correctly $6 \cdot 8 = 48$. But one of these must be negative, since we need the result to be $\textcolor{b l u e}{- 48}$. This brings us to the condition that the factors add to give $\textcolor{m a \ge n t a}{+ 2}$. This allows us to deduce that the positive factor must be the $8$ since $8 - 6 = 2$. These are our factors $a$ and $b$, therefore

$\left(x + 8\right) \left(x - 6\right) = 0$

Setting each factor equal to zero we can solve for $x$

$x + 8 = 0 \implies x = - 8$
$x - 6 = 0 \implies x = 6$

So the solutions are

$x = - 8 , + 6$

A quick graph of the function confirms our calculations:

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

Jul 24, 2016

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.