# 2.5 Quadratic equations  (Page 2/14)

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Given a quadratic equation with the leading coefficient of 1, factor it.

1. Find two numbers whose product equals c and whose sum equals b .
2. Use those numbers to write two factors of the form where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and $\text{\hspace{0.17em}}-2,$ the factors are $\text{\hspace{0.17em}}\left(x+1\right)\left(x-2\right).$
3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

## Solving a quadratic equation by factoring when the leading coefficient is not 1

Factor and solve the equation: $\text{\hspace{0.17em}}{x}^{2}+x-6=0.$

To factor $\text{\hspace{0.17em}}{x}^{2}+x-6=0,$ we look for two numbers whose product equals $\text{\hspace{0.17em}}-6\text{\hspace{0.17em}}$ and whose sum equals 1. Begin by looking at the possible factors of $\text{\hspace{0.17em}}-6.$

$\begin{array}{c}1\cdot \left(-6\right)\\ \left(-6\right)\cdot 1\\ 2\cdot \left(-3\right)\\ 3\cdot \left(-2\right)\end{array}$

The last pair, $\text{\hspace{0.17em}}3\cdot \left(-2\right)\text{\hspace{0.17em}}$ sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

$\left(x-2\right)\left(x+3\right)=0$

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

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

The two solutions are $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}-3.\text{\hspace{0.17em}}$ We can see how the solutions relate to the graph in [link] . The solutions are the x- intercepts of $\text{\hspace{0.17em}}y={x}^{2}+x-6=0.$

Factor and solve the quadratic equation: $\text{\hspace{0.17em}}{x}^{2}-5x-6=0.$

$\left(x-6\right)\left(x+1\right)=0;x=6,x=-1$

## Solve the quadratic equation by factoring

Solve the quadratic equation by factoring: $\text{\hspace{0.17em}}{x}^{2}+8x+15=0.$

Find two numbers whose product equals $\text{\hspace{0.17em}}15\text{\hspace{0.17em}}$ and whose sum equals $\text{\hspace{0.17em}}8.\text{\hspace{0.17em}}$ List the factors of $\text{\hspace{0.17em}}15.$

$\begin{array}{c}1\cdot 15\hfill \\ 3\cdot 5\hfill \\ \left(-1\right)\cdot \left(-15\right)\hfill \\ \left(-3\right)\cdot \left(-5\right)\hfill \end{array}$

The numbers that add to 8 are 3 and 5. Then, write the factors, set each factor equal to zero, and solve.

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

The solutions are $\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}-5.$

Solve the quadratic equation by factoring: $\text{\hspace{0.17em}}{x}^{2}-4x-21=0.$

$\left(x-7\right)\left(x+3\right)=0,$ $x=7,$ $x=-3.$

## Using the zero-product property to solve a quadratic equation written as the difference of squares

Solve the difference of squares equation using the zero-product property: $\text{\hspace{0.17em}}{x}^{2}-9=0.$

Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.

$\begin{array}{ccc}\hfill {x}^{2}-9& =& 0\hfill \\ \hfill \left(x-3\right)\left(x+3\right)& =& 0\hfill \\ \phantom{\rule{2em}{0ex}}\hfill \left(x-3\right)& =& 0\hfill \\ \hfill x& =& 3\hfill \\ \phantom{\rule{2em}{0ex}}\hfill \left(x+3\right)& =& 0\hfill \\ \hfill x& =& -3\hfill \end{array}$

The solutions are $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}-3.$

Solve by factoring: $\text{\hspace{0.17em}}{x}^{2}-25=0.$

$\left(x+5\right)\left(x-5\right)=0,$ $x=-5,$ $x=5.$

## Solving a quadratic equation by factoring when the leading coefficient is not 1

When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:

1. With the quadratic in standard form, $\text{\hspace{0.17em}}a{x}^{2}+bx+c=0,$ multiply $\text{\hspace{0.17em}}a\cdot c.$
2. Find two numbers whose product equals $\text{\hspace{0.17em}}ac\text{\hspace{0.17em}}$ and whose sum equals $\text{\hspace{0.17em}}b.$
3. Rewrite the equation replacing the $\text{\hspace{0.17em}}bx\text{\hspace{0.17em}}$ term with two terms using the numbers found in step 1 as coefficients of x.
4. Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
5. Factor out the expression in parentheses.
6. Set the expressions equal to zero and solve for the variable.

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
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Prove it
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I think you should say "28 terms" instead of "28th term"
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