Solving quadratic equations by factoring

Algebra ii for the community Page 1 / 1

This module is from Elementary Algebra</link>by Denny Burzynski and Wade Ellis, Jr. Methods of solving quadratic equations as well as the logic underlying each method are discussed. Factoring, extraction of roots, completing the square, and the quadratic formula are carefully developed. The zero-factor property of real numbers is reintroduced. The chapter also includes graphs of quadratic equations based on the standard parabola, y = x^2, and applied problems from the areas of manufacturing, population, physics, geometry, mathematics (numbers and volumes), and astronomy, which are solved using the five-step method.Objectives of this module: be able to solve quadratic equations by factoring.

Overview

Factoring Method
Solving Mentally After Factoring

Factoring method

To solve quadratic equations by factoring, we must make use of the zero-factor property.

Set the equation equal to zero, that is, get all the nonzero terms on one side of the equal sign and 0 on the other.

$a x^{2} + b x + c = 0$
Factor the quadratic expression.

$() () = 0$
By the zero-factor property, at least one of the factors must be zero, so, set each of the factors equal to 0 and solve for the variable.

Sample set a

Solve the following quadratic equations. (We will show the check for problem 1.)

$\begin{array}{l} x^{2} - 7 x + 12 & = & 0. & \begin{array}{l} The equation is already \\ set equal to 0 . Factor . \end{array} \\ (x - 3) (x - 4) & = & 0 & Set each factor equal to 0 . \\ x - 3 & = & 0 & or & x - 4 & = & 0 \\ x & = & 3 & or & x & = & 4 \end{array}$
$\begin{array}{l} C h e c k : If x = 3, x^{2} - 7 x & + & 12 & = & 0 \\ 3^{2} - 7 \cdot 3 & + & 12 & = & 0 & Is this correct? \\ 9 - 21 & + & 12 & = & 0 & Is this correct? \\ 0 & = & 0 & Yes, this is correct . \end{array}$

$\begin{array}{l} C h e c k : If x = 4, x^{2} - 7 x & + & 12 & = & 0 \\ 4^{2} - 7 \cdot 4 & + & 12 & = & 0 & Is this correct? \\ 16 - 28 & + & 12 & = & 0 & Is this correct? \\ 0 & = & 0 & Yes, this is correct . \end{array}$
Thus, the solutions to this equation are $x = 3, 4.$

$\begin{array}{l} x^{2} & = & 25. & Set the equation equal to 0. \\ x^{2} - 25 & = & 0 & Factor . \\ (x + 5) (x - 5) & = & 0 & Set each factor equal to 0. \\ x + 5 = 0 & or & x - 5 = 0 \\ x = - 5 & or & x = 5 \end{array}$
Thus, the solutions to this equation are $x = 5, - 5.$

$\begin{array}{l} x^{2} & = & 2 x . & Set the equation equal to 0. \\ x^{2} - 2 x & = & 0 & Factor . \\ x (x - 2) & Set each factor equal to 0. \\ x = 0 & or & x - 2 = 0 \\ x = 2 \end{array}$
Thus, the solutions to this equation are $x = 0, 2.$

$\begin{array}{l} 2 x^{2} + 7 x - 15 & = & 0. & Factor . \\ (2 x - 3) (x + 5) & = & 0 & Set each factor equal to 0. \\ 2 x - 3 = 0 & or & x + 5 = 0 \\ 2 x = 3 & or & x = - 5 \\ x = \frac{3}{2} \end{array}$
Thus, the solutions to this equation are $x = \frac{3}{2}, - 5.$

$63 x^{2} = 13 x + 6$
$\begin{array}{l} 63 x^{2} - 13 x - 6 & = & 0 \\ (9 x + 2) (7 x - 3) & = & 0 \\ 9 x + 2 = 0 & or & 7 x - 3 = 0 \\ 9 x = - 2 & or & 7 x = 3 \\ x = \frac{- 2}{9} & or & x = \frac{3}{7} \end{array}$
Thus, the solutions to this equation are $x = \frac{- 2}{9}, \frac{3}{7} .$

Practice set a

Solve the following equations, if possible.

$(x - 7) (x + 4) = 0$

$x = 7, - 4$

$(2 x + 5) (5 x - 7) = 0$

$x = \frac{- 5}{2}, \frac{7}{5}$

$x^{2} + 2 x - 24 = 0$

$x = 4, - 6$

$6 x^{2} + 13 x - 5 = 0$

$x = \frac{1}{3}, \frac{- 5}{2}$

$5 y^{2} + 2 y = 3$

$y = \frac{3}{5}, - 1$

$m (2 m - 11) = 0$

$m = 0, \frac{11}{2}$

$6 p^{2} = - (5 p + 1)$

$p = \frac{- 1}{3}, \frac{- 1}{2}$

$r^{2} - 49 = 0$

$r = 7, - 7$

Solving mentally after factoring

Let’s consider problems 4 and 5 of Sample Set A in more detail. Let’s look particularly at the factorizations $(2 x - 3) (x + 5) = 0$ and $(9 x + 2) (7 x - 3) = 0.$ The next step is to set each factor equal to zero and solve. We can solve mentally if we understand how to solve linear equations: we transpose the constant from the variable term and then divide by the coefficient of the variable.

Sample set b

Solve the following equation mentally.

$(2 x - 3) (x + 5) = 0$
$\begin{array}{l} 2 x - 3 & = & 0 & Mentally add 3 to both sides . The constant changes sign . \\ 2 x & = & 3 & \begin{array}{l} Divide by 2, the coefficient of x . The 2 divides the constant 3 into \frac{3}{2} . \\ The coefficient becomes the denominator . \end{array} \\ x & = & \frac{3}{2} \\ x + 5 & = & 0 & Mentally subtract 5 from both sides . The constant changes sign . \\ x & = & - 5 & Divide by the coefficient of x, 1 .The coefficient becomes the denominator . \\ x = \frac{- 5}{1} & = & - 5 \\ x & = & - 5 \end{array}$
Now, we can immediately write the solution to the equation after factoring by looking at each factor, changing the sign of the constant, then dividing by the coefficient.

Practice set b

Solve $(9 x + 2) (7 x - 3) = 0$ using this mental method.

$x = - \frac{2}{9}, \frac{3}{7}$

Exercises

For the following problems, solve the equations, if possible.

$(x + 1) (x + 3) = 0$

$x = - 1, - 3$

$(x + 4) (x + 9) = 0$

$(x - 5) (x - 1) = 0$

$x = 1, 5$

$(x - 6) (x - 3) = 0$

$(x - 4) (x + 2) = 0$

$x = - 2, 4$

$(x + 6) (x - 1) = 0$

$(2 x + 1) (x - 7) = 0$

$x = - \frac{1}{2}, 7$

$(3 x + 2) (x - 1) = 0$

$(4 x + 3) (3 x - 2) = 0$

$x = - \frac{3}{4}, \frac{2}{3}$

$(5 x - 1) (4 x + 7) = 0$

$(6 x + 5) (9 x - 4) = 0$

$x = - \frac{5}{6}, \frac{4}{9}$

$(3 a + 1) (3 a - 1) = 0$

$x (x + 4) = 0$

$x = - 4, 0$

$y (y - 5) = 0$

$y (3 y - 4) = 0$

$y = 0, \frac{4}{3}$

$b (4 b + 5) = 0$

$x (2 x + 1) (2 x + 8) = 0$

$x = - 4, - \frac{1}{2}, 0$

$y (5 y + 2) (2 y - 1) = 0$

${(x - 8)}^{2} = 0$

$x = 8$

${(x - 2)}^{2} = 0$

${(b + 7)}^{2} = 0$

$b = - 7$

${(a + 1)}^{2} = 0$

$x {(x - 4)}^{2} = 0$

$x = 0, 4$

$y {(y + 9)}^{2} = 0$

$y {(y - 7)}^{2} = 0$

$y = 0, 7$

$y {(y + 5)}^{2} = 0$

$x^{2} - 4 = 0$

$x = - 2, 2$

$x^{2} + 9 = 0$

$x^{2} + 36 = 0$

no solution

$x^{2} - 25 = 0$

$a^{2} - 100 = 0$

$a = - 10, 10$

$a^{2} - 81 = 0$

$b^{2} - 49 = 0$

$b = 7, - 7$

$y^{2} - 1 = 0$

$3 a^{2} - 75 = 0$

$a = 5, - 5$

$5 b^{2} - 20 = 0$

$y^{3} - y = 0$

$y = 0, 1, - 1$

$a^{2} = 9$

$b^{2} = 4$

$b = 2, - 2$

$b^{2} = 1$

$a^{2} = 36$

$a = 6, - 6$

$3 a^{2} = 12$

$- 2 x^{2} = - 4$

$x = \sqrt{2}, - \sqrt{2}$

$- 2 a^{2} = - 50$

$- 7 b^{2} = - 63$

$b = 3, - 3$

$- 2 x^{2} = - 32$

$3 b^{2} = 48$

$b = 4, - 4$

$a^{2} - 8 a + 16 = 0$

$y^{2} + 10 y + 25 = 0$

$y = - 5$

$y^{2} + 9 y + 16 = 0$

$x^{2} - 2 x - 1 = 0$

no solution

$a^{2} + 6 a + 9 = 0$

$a^{2} + 4 a + 4 = 0$

$a = - 2$

$x^{2} + 12 x = - 36$

$b^{2} - 14 b = - 49$

$b = 7$

$3 a^{2} + 18 a + 27 = 0$

$2 m^{3} + 4 m^{2} + 2 m = 0$

$m = 0, - 1$

$3 m n^{2} - 36 m n + 36 m = 0$

$a^{2} + 2 a - 3 = 0$

$a = - 3, 1$

$a^{2} + 3 a - 10 = 0$

$x^{2} + 9 x + 14 = 0$

$x = - 7, - 2$

$x^{2} - 7 x + 12 = 3$

$b^{2} + 12 b + 27 = 0$

$b = - 9, - 3$

$b^{2} - 3 b + 2 = 0$

$x^{2} - 13 x = - 42$

$x = 6, 7$

$a^{3} = - 8 a^{2} - 15 a$

$6 a^{2} + 13 a + 5 = 0$

$a = - \frac{5}{3}, - \frac{1}{2}$

$6 x^{2} - 4 x - 2 = 0$

$12 a^{2} + 15 a + 3 = 0$

$a = - \frac{1}{4}, - 1$

$18 b^{2} + 24 b + 6 = 0$

$12 a^{2} + 24 a + 12 = 0$

$a = - 1$

$4 x^{2} - 4 x = - 1$

$2 x^{2} = x + 15$

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

$4 a^{2} = 4 a + 3$

$4 y^{2} = - 4 y - 2$

no solution

$9 y^{2} = 9 y + 18$

Exercises for review

( [link] ) Simplify ${(x^{4} y^{3})}^{2} {(x y^{2})}^{4} .$

$x^{12} y^{14}$

( [link] ) Write ${(x^{- 2} y^{3} w^{4})}^{- 2}$ so that only positive exponents appear.

( [link] ) Find the sum: $\frac{x}{x^{2} - x - 2} + \frac{1}{x^{2} - 3 x + 2} .$

$\frac{x^{2} + 1}{(x + 1) (x - 1) (x - 2)}$

( [link] ) Simplify $\frac{\frac{1}{a} + \frac{1}{b}}{\frac{1}{a} - \frac{1}{b}} .$

( [link] ) Solve $(x + 4) (3 x + 1) = 0.$

$x = - 4, \frac{- 1}{3}$

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Source: OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1

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