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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Dakolo Reply

Communication is effective because it allows individuals to share ideas, thoughts, and information with others.

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Every time someone flushes a toilet in the apartment building, the person begins to jumb back automatically after hearing the flush, before the water temperature changes. Identify the types of learning, if it is classical conditioning identify the NS, UCS, CS and CR. If it is operant conditioning, identify the type of consequence positive reinforcement, negative reinforcement or punishment

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Wekolamo

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Skinner skipped the whole unconscious phenomenon and rather emphasized on classical conditioning

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nature is an hereditary factor while nurture is an environmental factor which constitute an individual personality. so if an individual's parent has a deviant behavior and was also brought up in an deviant environment, observation of the behavior and the inborn trait we make the individual deviant.

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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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