# 2.5 Quadratic equations  (Page 3/14)

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## Solving a quadratic equation using grouping

Use grouping to factor and solve the quadratic equation: $\text{\hspace{0.17em}}4{x}^{2}+15x+9=0.$

First, multiply $\text{\hspace{0.17em}}ac:4\left(9\right)=36.\text{\hspace{0.17em}}$ Then list the factors of $\text{\hspace{0.17em}}36.$

$\begin{array}{l}1\cdot 36\hfill \\ 2\cdot 18\hfill \\ 3\cdot 12\hfill \\ 4\cdot 9\hfill \\ 6\cdot 6\hfill \end{array}$

The only pair of factors that sums to $\text{\hspace{0.17em}}15\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}3+12.\text{\hspace{0.17em}}$ Rewrite the equation replacing the b term, $\text{\hspace{0.17em}}15x,$ with two terms using 3 and 12 as coefficients of x . Factor the first two terms, and then factor the last two terms.

$\begin{array}{ccc}\hfill 4{x}^{2}+3x+12x+9& =& 0\hfill \\ \hfill x\left(4x+3\right)+3\left(4x+3\right)& =& 0\hfill \\ \hfill \left(4x+3\right)\left(x+3\right)& =& 0\hfill \end{array}$

Solve using the zero-product property.

$\begin{array}{ccc}\hfill \left(4x+3\right)\left(x+3\right)& =& 0\hfill \\ \phantom{\rule{2em}{0ex}}\hfill \left(4x+3\right)& =& 0\hfill \\ \hfill x& =& -\frac{3}{4}\hfill \\ \phantom{\rule{2em}{0ex}}\hfill \left(x+3\right)& =& 0\hfill \\ \hfill x& =& -3\hfill \end{array}$

The solutions are $\text{\hspace{0.17em}}-\frac{3}{4},$ $\text{and}-3.\text{\hspace{0.17em}}$ See [link] .

Solve using factoring by grouping: $\text{\hspace{0.17em}}12{x}^{2}+11x+2=0.$

$\left(3x+2\right)\left(4x+1\right)=0,$ $x=-\frac{2}{3},$ $x=-\frac{1}{4}$

## Solving a polynomial of higher degree by factoring

Solve the equation by factoring: $\text{\hspace{0.17em}}-3{x}^{3}-5{x}^{2}-2x=0.$

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 $\text{\hspace{0.17em}}-x\text{\hspace{0.17em}}$ from all of the terms and then proceed with grouping.

$\begin{array}{ccc}\hfill -3{x}^{3}-5{x}^{2}-2x& =& 0\hfill \\ \hfill -x\left(3{x}^{2}+5x+2\right)& =& 0\hfill \end{array}$

Use grouping on the expression in parentheses.

$\begin{array}{ccc}\hfill -x\left(3{x}^{2}+3x+2x+2\right)& =& 0\hfill \\ \hfill -x\left[3x\left(x+1\right)+2\left(x+1\right)\right]& =& 0\hfill \\ \hfill -x\left(3x+2\right)\left(x+1\right)& =& 0\hfill \end{array}$

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

$\begin{array}{ccc}\hfill -x& =& 0\hfill \\ \hfill x& =& 0\hfill \\ \hfill 3x+2& =& 0\hfill \\ \hfill x& =& -\frac{2}{3}\hfill \\ \hfill x+1& =& 0\hfill \\ \hfill x& =& -1\hfill \end{array}$

The solutions are $\text{\hspace{0.17em}}0,$ $-\frac{2}{3},$ and $\text{\hspace{0.17em}}-1.$

Solve by factoring: $\text{\hspace{0.17em}}{x}^{3}+11{x}^{2}+10x=0.$

$x=0,x=-10,x=-1$

## Using the square root property

When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property    , in which we isolate the $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ term so that the square root property can be used.

## The square root property

With the $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ term isolated, the square root property states that:

$\text{if}\text{\hspace{0.17em}}{x}^{2}=k,\text{then}\text{\hspace{0.17em}}x=±\sqrt{k}$

where k is a nonzero real number.

Given a quadratic equation with an $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ term but no $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ term, use the square root property to solve it.

1. Isolate the $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ term on one side of the equal sign.
2. Take the square root of both sides of the equation, putting a $\text{\hspace{0.17em}}±\text{\hspace{0.17em}}$ sign before the expression on the side opposite the squared term.
3. Simplify the numbers on the side with the $\text{\hspace{0.17em}}±\text{\hspace{0.17em}}$ sign.

## Solving a simple quadratic equation using the square root property

Solve the quadratic using the square root property: $\text{\hspace{0.17em}}{x}^{2}=8.$

Take the square root of both sides, and then simplify the radical. Remember to use a $\text{\hspace{0.17em}}±\text{\hspace{0.17em}}$ sign before the radical symbol.

$\begin{array}{ccc}\hfill {x}^{2}& =& 8\hfill \\ \hfill x& =& ±\sqrt{8}\hfill \\ & =& ±2\sqrt{2}\hfill \end{array}$

The solutions are $\text{\hspace{0.17em}}2\sqrt{2},$ $-2\sqrt{2}.$

## Solving a quadratic equation using the square root property

Solve the quadratic equation: $\text{\hspace{0.17em}}4{x}^{2}+1=\text{7.}$

First, isolate the $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ term. Then take the square root of both sides.

$\begin{array}{ccc}\hfill 4{x}^{2}+1& =& 7\hfill \\ \hfill 4{x}^{2}& =& 6\hfill \\ \hfill {x}^{2}& =& \frac{6}{4}\hfill \\ \hfill x& =& ±\frac{\sqrt{6}}{2}\hfill \end{array}$

The solutions are $\text{\hspace{0.17em}}\frac{\sqrt{6}}{2},$ $\text{and}-\frac{\sqrt{6}}{2}.$

Solve the quadratic equation using the square root property: $\text{\hspace{0.17em}}3{\left(x-4\right)}^{2}=15.$

$x=4±\sqrt{5}$

## Completing the square

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation    known as completing the square    . Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a , must equal 1. If it does not, then divide the entire equation by a . Then, we can use the following procedures to solve a quadratic equation by completing the square.

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