Three Different Methods for Solving an Equation

9.4.4 Three Different Methods for Solving an Equation

9.4.4 • Three Different Methods for Solving an Equation

Activity

Here are three methods for solving $3 x^{2} + 8 x + 5 = 0$ .

Carefully examine each method.

Method 1:	Method 2:	Method 3:
$\begin{array}{rcl} 3 x^{2} + 8 x + 5 & = & 0 \\ (3 x + 5) (x + 1) & = & 0 \\ x + 1 & = & 0 \\ 3 x + 5 & = & 0 \\ x & = & - 1 \\ 3 x & = & - 5 \\ x & = & - \frac{5}{3} \\ \begin{array}{rcl} x & = & - \frac{5}{3} \end{array} & and & x = - 1 \end{array}$	$\begin{array}{rcl} 3 x^{2} + 8 x + 5 & = & 0 \\ 9 x^{2} + 24 x + 15 & = & 0 \\ (3 x)^{2} + 8 (3 x) + 15 & = & 0 \\ U^{2} + 8 U + 15 & = & 0 \\ (U + 5) (U + 3) & = & 0 \\ U + 5 & = & 0 \\ U + 3 & = & 0 \\ U & = & - 5 \\ U & = & - 3 \\ 3 x & = & - 5 \\ 3 x & = & - 3 \\ \begin{array}{rcl} x & = & - \frac{5}{3} \end{array} & and & x = - 1 \end{array}$	$\begin{array}{rcl} 3 x^{2} + 8 x + 5 & = & 0 \\ 9 x^{2} + 24 x + 15 & = & 0 \\ 9 x^{2} + 24 x + 16 & = & 1 \\ (3 x + 4)^{2} & = & 1 \\ 3 x + 4 & = & 1 \\ 3 x + 4 & = & - 1 \\ \begin{array}{rcl} x & = & - \frac{5}{3} \end{array} & and & x = - 1 \end{array}$

Once you understand the methods, use each method at least one time to solve these equations.

$5 x^{2} + 17 x + 6 = 0$

Compare your answer:

Method 1 is used here, but any of the three methods would work.

$\begin{matrix} 5 x^{2} + 17 x + 6 & = & 0 \\ (5 x + 2) (x + 3) & = & 0 \\ 5 x + 2 & = & 0 \\ 5 x & = & - 2 \\ x & = & - \frac{2}{5} \\ x + 3 & = & 0 \\ x & = & - 3 \end{matrix}$

$x = - \frac{2}{5} and x = - 3$

$6 x^{2} + 19 x = - 10$

Compare your answer:

Method 2 is used here, but any of the three methods would work.

$\begin{matrix} 6 x^{2} + 19 x & = & - 10 \\ 6 x^{2} + 19 x + 10 & = & 0 \\ 36 x^{2} + 114 x + 60 & = & 0 \\ (6 x)^{2} + 19 (6 x) + 60 & = & 0 \\ U^{2} + 19 U + 60 & = & 0 \\ (U + 15) (U + 4) & = & 0 \\ U + 15 & = & 0 \\ U & = & - 15 \\ 6 x & = & - 15 \\ x & = & - \frac{15}{6} = - \frac{5}{2} \\ (U + 4) & = & 0 \\ U & = & - 4 \\ 6 x & = & - 4 \\ x & = & - \frac{4}{6} = - \frac{2}{3} \end{matrix}$

$x = - \frac{5}{2} and x = - \frac{2}{3}$

$8 x^{2} - 33 x + 4 = 0$

Compare your answer:

Method 1 is used here, but any of the three methods would work.

$\begin{matrix} 8 x^{2} - 33 x + 4 & = & 0 \\ (8 x - 1) (x - 4) & = & 0 \\ 8 x - 1 & = & 0 \\ 8 x & = & 1 \\ x & = & \frac{1}{8} \\ x - 4 & = 0 \\ x & = & 4 \end{matrix}$

$x = \frac{1}{8} and x = 4$

$8 x^{2} - 26 x = - 21$

Compare your answer:

Method 1 is used here, but any of the three methods would work.

$\begin{matrix} 8 x^{2} - 26 x & = & - 21 \\ 8 x^{2} - 26 x + 21 & = & 0 \\ (4 x - 7) (2 x - 3) & = & 0 \\ 4 x - 7 & = & 0 \\ 4 x & = & 7 \\ x & = & \frac{7}{4} \\ 2 x - 3 & = & 0 \\ 2 x & = & 3 \\ x & = & \frac{3}{2} \\ x = \frac{7}{4} and x = \frac{3}{2} \end{matrix}$

$10 x^{2} + 37 x = 36$

Compare your answer:

Method 3 is used here, but any of the three methods would work.

$\begin{matrix} 10 x^{2} + 37 x & = & 36 \\ \frac{(10 x^{2} + 37 x)}{10} & = & \frac{36}{10} \\ x^{2} + \frac{37}{10} x & = & \frac{36}{10} \\ x^{2} + \frac{37}{10} x + ((\frac{1}{2}) (\frac{37}{10}))^{2} & = & \frac{36}{10} + ((\frac{1}{2}) (\frac{37}{10}))^{2} \\ x^{2} + \frac{37}{10} x + \frac{1369}{400} & = & \frac{36}{10} + \frac{1369}{400} \\ x^{2} + \frac{37}{10} x + \frac{1369}{400} & = & \frac{2809}{400} \\ (x + \frac{37}{20})^{2} & = & \frac{2809}{400} \\ \sqrt{(x + \frac{37}{20})^{2}} & = & \sqrt{\frac{2809}{400}} \\ x + \frac{37}{20} & = & \frac{53}{20} \\ x & = & \frac{16}{20} \\ x & = & \frac{4}{5} \\ x + \frac{37}{20} & = & - \frac{53}{20} \\ x & = & - \frac{90}{20} \\ x & = & - \frac{9}{2} \end{matrix}$

$x = \frac{4}{5} and x = - \frac{9}{2}$

$12 x^{2} + 20 x - 77 = 0$

Compare your answer:

Method 3 is used here, but any of the three methods would work.

$\begin{matrix} 12 x^{2} + 20 x - 77 & = & 0 \\ 12 x^{2} + 20 x & = & 77 \\ \frac{(12 x^{2} + 20 x)}{12} & = & \frac{77}{12} \\ x^{2} + \frac{20}{12} x & = & \frac{77}{12} \\ x^{2} + \frac{20}{12} x + {((\frac{1}{2}) (\frac{20}{12}))}^{2} & = & \frac{77}{12} + {((\frac{1}{2}) (\frac{20}{12}))}^{2} \\ x^{2} + \frac{20}{12} x + \frac{100}{144} & = & \frac{77}{12} + \frac{100}{144} = \frac{924}{144} + \frac{100}{144} \\ x^{2} + \frac{20}{12} x + \frac{100}{144} & = & \frac{1024}{144} \\ {(x + \frac{10}{12})}^{2} & = & \frac{1024}{144} \\ \sqrt{{(x + \frac{10}{12})}^{2}} & = & \sqrt{\frac{1024}{144}} \\ x + \frac{10}{12} & = & \frac{32}{12} \\ x + \frac{10}{12} & = & - \frac{32}{12} \\ x & = & \frac{22}{12} \\ x & = & - \frac{42}{12} \\ x & = & \frac{11}{6} \\ x & = & - \frac{21}{6} = - \frac{7}{2} \end{matrix}$

$x = \frac{11}{6} and x = - \frac{7}{2}$

Are you ready for more?

Extending Your Thinking

Solve the equation $2 x^{2} - 3 x + 1 = 0$ , then choose the two roots.

A. $x = - \frac{3}{2}$

B. $x = - \frac{5}{2}$

C. $x = 1$

D. $x = \frac{5}{2}$

E. $x = \frac{1}{2}$

Compare your answer:

C. $x = 1$ , E. $x = \frac{1}{2}$

Be prepared to show your reasoning for your solution.

Compare your answer:

Using the techniques of this lesson, the first step is to divide both sides of the equation by 2 to obtain $x^{2} - \frac{3}{2} x + \frac{1}{2} = 0$ , and then complete the square.