8.1 Quadratic equations: solving quadratic equations using the

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: recognize the standard form of a quadratic equation, understand the derivation of the quadratic formula, solve quadratic equations using the quadratic formula.

Overview

Standard Form Of A Quadratic Equation
The Quadratic Formula
Derivation Of The Quadratic Formula

Standard form of a quadratic equation

We have observed that a quadratic equation is an equation of the form

$\begin{array}{l} a x^{2} + b x + c = 0, & a \neq 0 \end{array}$

where

$a$ is the coefficient of the quadratic term,
$b$ is the coefficient of the linear term, and
$c$ is the constant term.

Standard form

The equation

a x^{2} + b x + c = 0

is the standard form of a quadratic equation.

Sample set a

Determine the values of $a, b,$ and $c .$

In the equation $3 x^{2} + 5 x + 2 = 0,$

$\begin{array}{l} a & = & 3 \\ b & = & 5 \\ c & = & 2 \end{array}$

In the equation $12 x^{2} - 2 x - 1 = 0,$

$\begin{array}{l} a & = & 12 \\ b & = & - 2 \\ c & = & - 1 \end{array}$

In the equation $2 y^{2} + 3 = 0,$

$\begin{array}{l} a & = & 2 \\ b & = & 0 & Because the equation could be written \\ 2 y^{2} + 0 y + 3 = 0 \\ c & = & 3 \end{array}$

In the equation $- 8 y^{2} + 11 y = 0,$

$\begin{array}{l} a & = & - 8 \\ b & = & 11 \\ c & = & 0 & Since - {8y}^{2} + 11 y + 0 = 0. \end{array}$

In the equation $z^{2} = z + 8,$

$\begin{array}{l} a & = & 1 \\ b & = & - 1 \\ c & = & - 8 & When we write the equation in standard form, we get z^{2} - z - 8 = 0 \end{array}$

Practice set a

Determine the values of $a, b,$ and $c$ in the following quadratic equations.

$4 x^{2} - 3 x + 5 = 0$

$\begin{array}{l} a & = & 4 \\ b & = & - 3 \\ c & = & 5 \end{array}$

$3 y^{2} - 2 y + 9 = 0$

$\begin{array}{l} a & = & 3 \\ b & = & - 2 \\ c & = & 9 \end{array}$

$x^{2} - 5 x - 1 = 0$

$\begin{array}{l} a & = & 1 \\ b & = & - 5 \\ c & = & - 1 \end{array}$

$z^{2} - 4 = 0$

$\begin{array}{l} a & = & 1 \\ b & = & 0 \\ c & = & - 4 \end{array}$

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

$\begin{array}{l} a & = & 1 \\ b & = & - 2 \\ c & = & 0 \end{array}$

$y^{2} = 5 y - 6$

$\begin{array}{l} a & = & 1 \\ b & = & - 5 \\ c & = & 6 \end{array}$

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

$\begin{array}{l} a & = & 2 \\ b & = & - 4 \\ c & = & 1 \end{array}$

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

$\begin{array}{l} a & = & 3 \\ b & = & 5 \\ c & = & - 3 \end{array}$

$2 x - 11 - 3 x^{2} = 0$

$\begin{array}{l} a & = & - 3 \\ b & = & 2 \\ c & = & - 11 \end{array}$

$y^{2} = 0$

$\begin{array}{l} a & = & 1 \\ b & = & 0 \\ c & = & 0 \end{array}$

The solutions to all quadratic equations depend only and completely on the values $a, b,$ and $c .$

The quadratic formula

When a quadratic equation is written in standard form so that the values $a, b,$ and $c$ are readily determined, the equation can be solved using the quadratic formula . The values that satisfy the equation are found by substituting the values $a, b,$ and $c$ into the formula

Quadratic formula

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Keep in mind that the plus or minus symbol, $\pm,$ is just a shorthand way of denoting the two possibilities:

$\begin{array}{l} x = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} & and & x = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array}$

The quadratic formula can be derived by using the method of completing the square.

Derivation of the quadratic formula

Solve $a x^{2} + b x + c = 0$ for $x$ by completing the square.

Subtract $c$ from both sides.

$a x^{2} + b x = - c$

Divide both sides by $a,$ the coefficient of $x^{2} .$

$x^{2} + \frac{b}{a} x = \frac{- c}{a}$

Now we have the proper form to complete the square. Take one half the coefficient of $x,$ square it, and add the result to both sides of the equation found in step 2.

(a) $\frac{1}{2} \cdot \frac{b}{a} = \frac{b}{2 a}$ is one half the coefficient of $x .$

(b) ${(\frac{b}{2 a})}^{2}$ is the square of one half the coefficient of $x .$

$x^{2} + \frac{b}{a} x + {(\frac{b}{2 a})}^{2} = \frac{- c}{a} + {(\frac{b}{2 a})}^{2}$

The left side of the equation is now a perfect square trinomial and can be factored. This gives us

${(x + \frac{b}{2 a})}^{2} = \frac{- c}{a} + \frac{b^{2}}{4 a^{2}}$

Add the two fractions on the right side of the equation. The LCD $= 4 a^{2} .$

$\begin{array}{l} {(x + \frac{b}{2 a})}^{2} & = & \frac{- 4 a c}{4 a^{2}} + \frac{b^{2}}{4 a^{2}} \\ {(x + \frac{b}{2 a})}^{2} & = & \frac{- 4 a c + b^{2}}{4 a^{2}} \\ {(x + \frac{b}{2 a})}^{2} & = & \frac{b^{2} - 4 a c}{4 a^{2}} \end{array}$

Solve for $x$ using the method of extraction of roots.

$\begin{array}{l} x + \frac{b}{2 a} & = & \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} \\ x + \frac{b}{2 a} & = & \pm \frac{\sqrt{b^{2} - 4 a c}}{\sqrt{4 a^{2}}} & \sqrt{4 a^{2}} = | 2 a | = 2 | a | = \pm 2 a \\ x + \frac{b}{2 a} & = & \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} \\ x & = & - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} & Add these two fractions . \\ x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{array}$

Sample set b

Solve each of the following quadratic equations using the quadratic formula.

$3 x^{2} + 5 x + 2 = 0.$

Identify $a, b,$ and $c .$

$\begin{array}{l} a = 3, & b = 5, & and & c = 2 \end{array}$
Write the quadratic formula.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Substitute.

$\begin{array}{l} x & = & \frac{- 5 \pm \sqrt{{(5)}^{2} - 4 (3) (2)}}{2 (3)} \\ = & \frac{- 5 \pm \sqrt{25 - 24}}{6} \\ = & \frac{- 5 \pm \sqrt{1}}{6} \\ = & \frac{- 5 \pm 1}{6} & - 5 + 1 = - 4 and - 5 - 1 = - 6 \\ = & \frac{- 4}{6}, \frac{- 6}{6} \\ x & = & \frac{- 2}{3}, - 1 \end{array}$
$\begin{array}{l} N o t e : & Since these roots are rational numbers, this equation could have been solved by factoring . \end{array}$

$12 x^{2} - 2 x - 1 = 0.$

Identify $a, b,$ and $c .$

$\begin{array}{l} a = 12, & b = - 2, & and & c = - 1 \end{array}$
Write the quadratic formula.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Substitute.

$\begin{array}{l} x & = & \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (12) (- 1)}}{2 (12)} \\ = & \frac{2 \pm \sqrt{4 + 48}}{24} & Simplify . \\ = & \frac{2 \pm \sqrt{52}}{24} & Simplify . \\ = & \frac{2 \pm \sqrt{4 \cdot 13}}{24} & Simplify . \\ = & \frac{2 \pm 2 \sqrt{13}}{24} & Reduce . Factor 2 from the terms of the numerator . \\ = & \frac{2 (1 \pm \sqrt{13})}{24} \\ x & = & \frac{1 \pm \sqrt{13}}{12} \end{array}$

$2 y^{2} + 3 = 0$

Identify $a, b,$ and $c .$

$\begin{array}{l} a = 2, & b = 0, & and & c = 3 \end{array}$
Write the quadratic formula.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Substitute.

$\begin{array}{l} x & = & \frac{- 0 \pm \sqrt{0^{2} - 4 (2) (3)}}{2 (2)} \\ x & = & \frac{0 \pm \sqrt{- 24}}{4} \end{array}$

This equation has no real number solution since we have obtained a negative number under the radical sign.

$- 8 x^{2} + 11 x = 0$

Identify $a, b,$ and $c .$

$\begin{array}{l} a = - 8, & b = 11, & and & c = 0 \end{array}$
Write the quadratic formula.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Substitute.

$\begin{array}{l} x & = & \frac{- 11 \pm \sqrt{11^{2} - 4 (- 8) (0)}}{2 (- 8)} \\ = & \frac{- 11 \pm \sqrt{121 - 0}}{- 16} & Simplify . \\ = & \frac{- 11 \pm \sqrt{121}}{- 16} & Simplify . \\ = & \frac{- 11 \pm 11}{- 16} \\ x & = & 0, \frac{11}{8} \end{array}$

$(3 x + 1) (x - 4) = x^{2} + x - 2$

Write the equation in standard form.

$\begin{array}{l} 3 x^{2} - 11 x - 4 & = & x^{2} + x - 2 \\ 2 x^{2} - 12 x - 2 & = & 0 \\ x^{2} - 6 x - 1 & = & 0 \end{array}$
Identify $a, b,$ and $c .$

$\begin{array}{l} a = 1, & b = - 6, & and & c = - 1 \end{array}$
Write the quadratic formula.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Substitute.

$\begin{array}{l} x & = & \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 (1) (- 1)}}{2 (1)} \\ = & \frac{6 \pm \sqrt{36 + 4}}{2} \\ = & \frac{6 \pm \sqrt{40}}{2} \\ = & \frac{6 \pm \sqrt{4 \cdot 10}}{2} \\ = & \frac{6 \pm 2 \sqrt{10}}{2} \\ = & \frac{2 (3 \pm \sqrt{10})}{2} \\ x & = & 3 \pm \sqrt{10} \end{array}$

Practice set b

Solve each of the following quadratic equations using the quadratic formula.

$2 x^{2} + 3 x - 7 = 0$

$x = \frac{- 3 \pm \sqrt{65}}{4}$

$5 a^{2} - 2 a - 1 = 0$

$a = \frac{1 \pm \sqrt{6}}{5}$

$6 y^{2} + 5 = 0$

no real number solution

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

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

Exercises

For the following problems, solve the equations using the quadratic formula.

$x^{2} - 2 x - 3 = 0$

$x = 3, - 1$

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

$y^{2} - 5 y + 4 = 0$

$y = 1, 4$

$a^{2} + 4 a - 21 = 0$

$a^{2} + 12 a + 20 = 0$

$a = - 2, - 10$

$b^{2} - 4 b + 4 = 0$

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

$b = - 2$

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

$2 x^{2} - 5 x - 3 = 0$

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

$6 y^{2} + y - 2 = 0$

$4 x^{2} - 2 x - 1 = 0$

$x = \frac{1 \pm \sqrt{5}}{4}$

$3 y^{2} + 2 y - 1 = 0$

$5 a^{2} - 2 a - 3 = 0$

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

$x^{2} - 3 x + 1 = 0$

$x^{2} - 5 x - 4 = 0$

$x = \frac{5 \pm \sqrt{41}}{2}$

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

$(a + 4) (a - 5) = 2$

$a = \frac{1 \pm \sqrt{89}}{2}$

$(x - 3) (x + 3) = 7$

$(b - 4) (b + 4) = 9$

$b = \pm 5$

$x^{2} + 8 x = 2$

$y^{2} = - 5 y + 4$

$y = \frac{- 5 \pm \sqrt{41}}{2}$

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

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

$x = - 1$

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

$a^{2} + 3 a - 4 = 0$

$a = - 4, 1$

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

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

$b = - 1, - 2$

$x^{2} + 6 x + 8 = - x - 2$

$x^{2} + 4 x = 2 x - 5$

No real number solution.

$6 b^{2} + 5 b - 4 = b^{2} + b + 1$

$4 a^{2} + 7 a - 2 = - 2 a + a$

$\frac{- 2 \pm \sqrt{6}}{2}$

$(2 x + 5) (x - 4) = x^{2} - x + 2$

${(x - 4)}^{2} = 3$

$x = 4 \pm \sqrt{3}$

${(x + 2)}^{2} = 4$

${(b - 6)}^{2} = 8$

$b = 6 \pm 2 \sqrt{2}$

${(3 - x)}^{2} = 6$

$3 (x^{2} + 1) = 2 (x + 7)$

$x = \frac{1 \pm \sqrt{34}}{3}$

$2 (y^{2} - 3) = - 3 (y - 1)$

$- 4 (a^{2} + 2) + 3 = 5$

No real number solution.

$- (x^{2} + 3 x - 1) = 2$

Exercises for review

( [link] ) Simplify ${(\frac{x^{8} y^{7} z^{5}}{x^{4} y^{6} z^{2}})}^{2} .$

$x^{8} y^{2} z^{6}$

( [link] ) Write $4 a^{- 6} b^{2} c^{3} a^{5} b^{- 3}$ so that only positive exponents appear.

( [link] ) Find the product: $(2 y + 7) (3 y - 1) .$

$6 y^{2} + 19 y - 7$

( [link] ) Simplify: $\sqrt{80} - \sqrt{45} .$

( [link] ) Solve $x^{2} - 4 x - 12 = 0$ by completing the square.

$x = - 2, 6$

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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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