# 2.5 Quadratic equations  (Page 4/14)

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We will use the example $\text{\hspace{0.17em}}{x}^{2}+4x+1=0\text{\hspace{0.17em}}$ to illustrate each step.

1. Given a quadratic equation that cannot be factored, and with $\text{\hspace{0.17em}}a=1,$ first add or subtract the constant term to the right sign of the equal sign.

${x}^{2}+4x=-1$
2. Multiply the b term by $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ and square it.

$\begin{array}{ccc}\hfill \frac{1}{2}\left(4\right)& =& 2\hfill \\ \hfill {2}^{2}& =& 4\hfill \end{array}$
3. Add $\text{\hspace{0.17em}}{\left(\frac{1}{2}b\right)}^{2}\text{\hspace{0.17em}}$ to both sides of the equal sign and simplify the right side. We have

$\begin{array}{ccc}\hfill {x}^{2}+4x+4& =& -1+4\hfill \\ \hfill {x}^{2}+4x+4& =& 3\hfill \end{array}$
4. The left side of the equation can now be factored as a perfect square.

$\begin{array}{ccc}\hfill {x}^{2}+4x+4& =& 3\hfill \\ \hfill {\left(x+2\right)}^{2}& =& 3\hfill \end{array}$
5. Use the square root property and solve.

$\begin{array}{ccc}\hfill \sqrt{{\left(x+2\right)}^{2}}& =& ±\sqrt{3}\hfill \\ \hfill x+2& =& ±\sqrt{3}\hfill \\ \hfill x& =& -2±\sqrt{3}\hfill \end{array}$
6. The solutions are $\text{\hspace{0.17em}}-2+\sqrt{3},$ $\text{and}-2-\sqrt{3}.$

## Solving a quadratic by completing the square

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

First, move the constant term to the right side of the equal sign.

${x}^{2}-3x=5$

Then, take $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ of the b term and square it.

$\begin{array}{ccc}\hfill \frac{1}{2}\left(-3\right)& =& -\frac{3}{2}\hfill \\ \hfill {\left(-\frac{3}{2}\right)}^{2}& =& \frac{9}{4}\hfill \end{array}$

Add the result to both sides of the equal sign.

$\begin{array}{ccc}\hfill {x}^{2}-3x+{\left(-\frac{3}{2}\right)}^{2}& =& 5+{\left(-\frac{3}{2}\right)}^{2}\hfill \\ \hfill {x}^{2}-3x+\frac{9}{4}& =& 5+\frac{9}{4}\hfill \end{array}$

Factor the left side as a perfect square and simplify the right side.

${\left(x-\frac{3}{2}\right)}^{2}=\frac{29}{4}$

Use the square root property and solve.

$\begin{array}{ccc}\hfill \sqrt{{\left(x-\frac{3}{2}\right)}^{2}}& =& ±\sqrt{\frac{29}{4}}\hfill \\ \hfill \left(x-\frac{3}{2}\right)& =& ±\frac{\sqrt{29}}{2}\hfill \\ \hfill x& =& \frac{3}{2}±\frac{\sqrt{29}}{2}\hfill \end{array}$

The solutions are $\text{\hspace{0.17em}}\frac{3}{2}+\frac{\sqrt{29}}{2},$ $\text{and}\frac{3}{2}-\frac{\sqrt{29}}{2}.$

Solve by completing the square: $\text{\hspace{0.17em}}{x}^{2}-6x=13.$

$x=3±\sqrt{22}$

The fourth method of solving a quadratic equation    is by using the quadratic formula    , a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.

We can derive the quadratic formula by completing the square    . We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by $\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}$ and obtain a positive a . Given $\text{\hspace{0.17em}}a{x}^{2}+bx+c=0,$ $a\ne 0,$ we will complete the square as follows:

1. First, move the constant term to the right side of the equal sign:

$a{x}^{2}+bx=-c$
2. As we want the leading coefficient to equal 1, divide through by a :

${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$
3. Then, find $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ of the middle term, and add $\text{\hspace{0.17em}}{\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}\text{\hspace{0.17em}}$ to both sides of the equal sign:

${x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}$
4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:

${\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$
5. Now, use the square root property, which gives

$\begin{array}{ccc}\hfill x+\frac{b}{2a}& =& ±\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ \hfill x+\frac{b}{2a}& =& \frac{±\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
6. Finally, add $\text{\hspace{0.17em}}-\frac{b}{2a}\text{\hspace{0.17em}}$ to both sides of the equation and combine the terms on the right side. Thus,

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

Written in standard form, $\text{\hspace{0.17em}}a{x}^{2}+bx+c=0,$ any quadratic equation can be solved using the quadratic formula    :

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

where a , b , and c are real numbers and $\text{\hspace{0.17em}}a\ne 0.$

1. Make sure the equation is in standard form: $\text{\hspace{0.17em}}a{x}^{2}+bx+c=0.$
2. Make note of the values of the coefficients and constant term, $\text{\hspace{0.17em}}a,b,$ and $\text{\hspace{0.17em}}c.$
3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
4. Calculate and solve.

Solve the quadratic equation: $\text{\hspace{0.17em}}{x}^{2}+5x+1=0.$

Identify the coefficients: $\text{\hspace{0.17em}}a=1,b=5,c=1.\text{\hspace{0.17em}}$ Then use the quadratic formula.

$\begin{array}{ccc}\hfill x& =& \hfill \frac{-\left(5\right)±\sqrt{{\left(5\right)}^{2}-4\left(1\right)\left(1\right)}}{2\left(1\right)}\\ & =& \frac{-5±\sqrt{25-4}}{2}\hfill \\ & =& \frac{-5±\sqrt{21}}{2}\hfill \end{array}$

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