2.5 Quadratic equations  (Page 3/14)

First, multiply $ac:4\left(9\right)=36.$ Then list the factors of $36.$

The only pair of factors that sums to $15$ is $3+12.$ Rewrite the equation replacing the b term, $15x,$ with two terms using 3 and 12 as coefficients of x . Factor the first two terms, and then factor the last two terms.

Solve using the zero-product property.

The solutions are $-\frac{3}{4},$ $\text{and}\mathrm{-3.}$ See [link] .

This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out $-x$ from all of the terms and then proceed with grouping.

Use grouping on the expression in parentheses.

Now, we use the zero-product property. Notice that we have three factors.

The solutions are $0,$ $-\frac{2}{3},$ and $\mathrm{-1.}$

Take the square root of both sides, and then simplify the radical. Remember to use a $\pm$ sign before the radical symbol.

The solutions are $2\sqrt{2},$ $\mathrm{-2}\sqrt{2}.$

First, isolate the $x^2$ term. Then take the square root of both sides.

The solutions are $\frac{\sqrt{6}}{2},$ $\text{and}-\frac{\sqrt{6}}{2}.$