7.6 Quadratic equations (Page 3/9)

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Solve: $6 a^{2} + 9 a = 3 a$ .

$a = 0, a = −1$

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Solve: $45 b^{2} - 2 b = −17 b$ .

$b = 0, b = - \frac{1}{3}$

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Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

Solve: $144 q^{2} = 25$ .

Solution

$\begin{matrix} \begin{matrix} Write the quadratic equation in standard form. \\ Factor. It is a difference of squares. \end{matrix} & \begin{matrix} 144 q^{2} & = & 25 \\ 144 q^{2} - 25 & = & 0 \\ (12 q - 5) (12 q + 5) & = & 0 \end{matrix} \\ \begin{matrix} Use the Zero Product Property to set each factor to 0 . \\ Solve each equation. \end{matrix} & \begin{matrix} 12 q - 5 & = & 0 & 12 q + 5 & = & 0 \\ 12 q & = & 5 & 12 q & = & −5 \\ q & = & \frac{5}{12} & q & = & - \frac{5}{12} \end{matrix} \\ Check your answers. \end{matrix}$

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Solve: $25 p^{2} = 49$ .

$p = \frac{7}{5}, p = - \frac{7}{5}$

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Solve: $36 x^{2} = 121$ .

$x = \frac{11}{6}, x = - \frac{11}{6}$

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The left side in the next example is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

Solve: $(3 x - 8) (x - 1) = 3 x$ .

Solution

$\begin{matrix} \begin{matrix} Multiply the binomials. \\ Write the quadratic equation in standard form. \\ Factor the trinomial. \end{matrix} & \begin{matrix} (3 x - 8) (x - 1) & = & 3 x \\ 3 x^{2} - 11 x + 8 & = & 3 x \\ 3 x^{2} - 14 x + 8 & = & 0 \\ (3 x - 2) (x - 4) & = & 0 \end{matrix} \\ \begin{matrix} Use the Zero Product Property to set each factor to 0. \\ Solve each equation. \end{matrix} & \begin{matrix} 3 x - 2 & = & 0 & x - 4 & = & 0 \\ 3 x & = & 2 & x & = & 4 \\ x & = & \frac{2}{3} \end{matrix} \\ Check your answers. & The check is left to you! \end{matrix}$

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Solve: $(2 m + 1) (m + 3) = 12 m$ .

$m = 1, m = \frac{3}{2}$

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Solve: $(k + 1) (k - 1) = 8$ .

$k = 3, k = −3$

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The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree more than two by using the Zero Product Property, just like we solved quadratic equations.

Solve: $9 m^{3} + 100 m = 60 m^{2}$ .

Solution

$\begin{matrix} \begin{matrix} Bring all the terms to one side so that the other side is zero. \\ Factor the greatest common factor first. \\ Factor the trinomial. \end{matrix} & \begin{matrix} 9 m^{3} + 100 m & = & 60 m^{2} \\ 9 m^{3} - 60 m^{2} + 100 m & = & 0 \\ m (9 m^{2} - 60 m + 100) & = & 0 \\ m (3 m - 10) (3 m - 10) & = & 0 \end{matrix} \\ \begin{matrix} Use the Zero Product Property to set each factor to 0 . \\ Solve each equation. \end{matrix} & \begin{matrix} m & = & 0 & 3 m - 10 & = & 0 & 3 m - 10 & = & 0 \\ m & = & 0 & m & = & \frac{10}{3} & m & = & \frac{10}{3} \end{matrix} \\ Check your answers. & The check is left to you. \end{matrix}$

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Solve: $8 x^{3} = 24 x^{2} - 18 x$ .

$x = 0, x = \frac{3}{2}$

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Solve: $16 y^{2} = 32 y^{3} + 2 y$ .

$y = 0, y = \frac{1}{4}$

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When we factor the quadratic equation in the next example we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

Solve: $4 x^{2} = 16 x + 84$ .

Solution

$\begin{matrix} \begin{matrix} Write the quadratic equation in standard form. \\ Factor the greatest common factor first. \\ Factor the trinomial. \end{matrix} & \begin{matrix} 4 x^{2} & = & 16 x + 84 \\ 4 x^{2} - 16 x - 84 & = & 0 \\ 4 (x^{2} - 4 x - 21) & = & 0 \\ 4 (x - 7) (x + 3) & = & 0 \end{matrix} \\ \begin{matrix} Use the Zero Product Property to set each factor to 0. \\ Solve each equation. \end{matrix} & \begin{matrix} 4 & \neq & 0 & x - 7 & = & 0 & x + 3 & = & 0 \\ 4 & \neq & 0 & x & = & 7 & x & = & −3 \end{matrix} \\ Check your answers. & The check is left to you. \end{matrix}$

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Solve: $18 a^{2} - 30 = −33 a$ .

$a = - \frac{5}{2}, a = \frac{2}{3}$

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Solve: $123 b = −6 - 60 b^{2}$ .

$b = 2, b = \frac{1}{20}$

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Solve applications modeled by quadratic equations

The problem solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to quadratic equations. We will copy the problem solving strategy here so we can use it for reference.

Use a problem-solving strategy to solve word problems.

Read the problem. Make sure all the words and ideas are understood.
Identify what we are looking for.
Name what we are looking for. Choose a variable to represent that quantity.
Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

Questions & Answers

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Not really sure

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The lymphatic system plays several crucial roles in the human body, functioning as a key component of the immune system and contributing to the maintenance of fluid balance. Its main functions include: 1. Immune Response: The lymphatic system produces and transports lymphocytes, which are a type of

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Anatomy is the identification and description of the structures of living things

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Anatomy is the study of the structure of the body, while physiology is the study of the function of the body. Anatomy looks at the body's organs and systems, while physiology looks at how those organs and systems work together to keep the body functioning.

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Enzymes are proteins that help speed up chemical reactions in our bodies. Enzymes are essential for digestion, liver function and much more. Too much or too little of a certain enzyme can cause health problems

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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