Quadratic equations: solving quadratic equations using the method

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 using the method of extraction of roots, be able to determine the nature of the solutions to a quadratic equation.

Overview

The Method Of Extraction Of Roots
The Nature Of Solutions

The method of extraction of roots

Extraction of roots

Quadratic equations of the form

x^{2} - K = 0

can be solved by the method of extraction of roots by rewriting it in the form

x^{2} = K .

To solve

x^{2} = K,

we are required to find some number,

x,

that when squared produces

K .

This number,

x,

must be a square root of

K .

K

is greater than zero, we know that it possesses two square roots,

\sqrt{K}

and

- \sqrt{K} .

We also know that

\begin{matrix} {(\sqrt{K})}^{2} = (\sqrt{K}) (\sqrt{K}) = K & and & {(- \sqrt{K})}^{2} = (- \sqrt{K}) (- \sqrt{K}) \end{matrix} = K

We now have two replacements for

x

that produce true statements when substituted into the equation. Thus,

x = \sqrt{K}

and

x = - \sqrt{K}

are both solutions to

x^{2} = K .

We use the notation

x = \pm \sqrt{K}

to denote both the principal and the secondary square roots.

The nature of solutions

Solutions of $x^{2} = K$

For quadratic equations of the form

x^{2} = K,

If $K$ is greater than or equal to zero, the solutions are $\pm \sqrt{K} .$
If $K$ is negative, no real number solutions exist.
If $K$ is zero, the only solution is 0.

Sample set a

Solve each of the following quadratic equations using the method of extraction of roots.

$\begin{matrix} x^{2} - 49 & = & 0. & Rewrite . \\ x^{2} & = & 49 \\ x & = & \pm \sqrt{49} \\ x & = & \pm 7 \\ C h e c k : & {(7)}^{2} = 49 & Is this correct? & {(- 7)}^{2} = 49 & Is this correct \\ 49 = 49 & Yes, this is correct . & 49 = 49 & Yes, this is correct . \end{matrix}$

$\begin{array}{r} 25 a^{2} & = & 36 \\ a^{2} & = & \frac{36}{25} \\ a & = & \pm \sqrt{\frac{36}{25}} \\ a & = & \pm \frac{6}{5} \end{array}$
$\begin{array}{l} C h e c k : & 25 {(\frac{6}{5})}^{2} & = & 36 & Is this correct? & 25 {(\frac{- 6}{5})}^{2} & = & 36 & Is this correct? \\ 25 {(\frac{36}{25})}^{2} & = & 36 & Is this correct? & 25 (\frac{36}{25}) & = & 36 & Is this correct? \\ 36 & = & 36 & Yes, this is correct . & 36 & = & 36 & Yes, this is correct . \end{array}$

$\begin{array}{r} 4 m^{2} - 32 & = & 0 \\ 4 m^{2} & = & 32 \\ m^{2} & = & \frac{32}{4} \\ m^{2} & = & 8 \\ m & = & \pm \sqrt{8} \\ m & = & \pm 2 \sqrt{2} \end{array}$
$\begin{array}{l} C h e c k : & 4 {(2 \sqrt{2})}^{2} & = & 32 & Is this correct? & 4 {(- 2 \sqrt{2})}^{2} & = & 32 & Is this correct? \\ 4 [2^{2} {(\sqrt{2})}^{2}] & = & 32 & Is this correct? & 4 [{(- 2)}^{2} {(\sqrt{2})}^{2}] & = & 32 & Is this correct? \\ 4 [4 \cdot 2] & = & 32 & Is this correct? & 4 [4 \cdot 2] & = & 32 & Is this correct? \\ 4 \cdot 8 & = & 32 & Is this correct? & 4 \cdot 8 & = & 32 & Is this correct? \\ 32 & = & 32 & Yes, this is correct . & 32 & = & 32 & Yes, this is correct . \end{array}$

Solve $5 x^{2} - 15 y^{2} z^{7} = 0$ for $x .$
$\begin{array}{l} 5 x^{2} & = & 15 y^{2} z^{7} & Divide both sides by 5 . \\ x^{2} & = & 3 y^{2} z^{7} \\ x & = & \pm \sqrt{3 y^{2} z^{7}} \\ x & = & \pm y z^{3} \sqrt{3 z} \end{array}$

Calculator problem. Solve $14 a^{2} - 235 = 0.$ Round to the nearest hundredth.
$\begin{matrix} 14 a^{2} - 235 & = & 0. & Rewrite . \\ 14 a^{2} & = & 235 & Divide both sides by 14 . \\ a^{2} & = & \frac{235}{14} \end{matrix}$
$On the Calculator$
$\begin{matrix} Type & 235 \\ Press & \begin{matrix} \div \end{matrix} \\ Type & 14 \\ Press & \begin{matrix} = \end{matrix} \\ Press & \sqrt \\ Display reads: & 4.0970373 \end{matrix}$
Rounding to the nearest hundredth produces 4.10. We must be sure to insert the $\pm$ symbol. $a \approx \pm 4.10$

$\begin{array}{l} k^{2} & = & - 64 \\ k & = & \pm \sqrt{- 64} \end{array}$
The radicand is negative so no real number solutions exist.

Practice set a

Solve each of the following quadratic equations using the method of extraction of roots.

$x^{2} - 144 = 0$

$x = \pm 12$

$9 y^{2} - 121 = 0$

$y = \pm \frac{11}{3}$

$6 a^{2} = 108$

$a = \pm 3 \sqrt{2}$

Solve $4 n^{2} = 24 m^{2} p^{8}$ for $n .$

$n = \pm m p^{4} \sqrt{6}$

Solve $5 p^{2} q^{2} = 45 p^{2}$ for $q .$

$q = \pm 3$

Solve $16 m^{2} - 2206 = 0.$ Round to the nearest hundredth.

$m = \pm 11.74$

$h^{2} = - 100$

Sample set b

Solve each of the following quadratic equations using the method of extraction of roots.

$\begin{array}{l} {(x + 2)}^{2} & = & 81 \\ x + 2 & = & \pm \sqrt{81} \\ x + 2 & = & \pm 9 & Subtract 2 from both sides . \\ x & = & - 2 \pm 9 \\ x & = & - 2 + 9 & and & x & = & - 2 - 9 \\ x & = & 7 & x & = & - 11 \end{array}$

$\begin{array}{l} {(a + 3)}^{2} & = & 5 \\ a + 3 & = & \pm \sqrt{5} & Subtract 3 from both sides . \\ a & = & - 3 \pm \sqrt{5} \end{array}$

Practice set b

Solve each of the following quadratic equations using the method of extraction of roots.

${(a + 6)}^{2} = 64$

$a = 2, - 14$

${(m - 4)}^{2} = 15$

$m = 4 \pm \sqrt{15}$

${(y - 7)}^{2} = 49$

$y = 0, 14$

${(k - 1)}^{2} = 12$

$k = 1 \pm 2 \sqrt{3}$

${(x - 11)}^{2} = 0$

$x = 11$

Exercises

For the following problems, solve each of the quadratic equations using the method of extraction of roots.

$x^{2} = 36$

$x = \pm 6$

$x^{2} = 49$

$a^{2} = 9$

$a = \pm 3$

$a^{2} = 4$

$b^{2} = 1$

$b = \pm 1$

$a^{2} = 1$

$x^{2} = 25$

$x = \pm 5$

$x^{2} = 81$

$a^{2} = 5$

$a = \pm \sqrt{5}$

$a^{2} = 10$

$b^{2} = 12$

$b = \pm 2 \sqrt{3}$

$b^{2} = 6$

$y^{2} = 3$

$y = \pm \sqrt{3}$

$y^{2} = 7$

$a^{2} - 8 = 0$

$a = \pm 2 \sqrt{2}$

$a^{2} - 3 = 0$

$a^{2} - 5 = 0$

$a = \pm \sqrt{5}$

$y^{2} - 1 = 0$

$x^{2} - 10 = 0$

$x = \pm \sqrt{10}$

$x^{2} - 11 = 0$

$3 x^{2} - 27 = 0$

$x = \pm 3$

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

$2 x^{2} = 50$

$x = \pm 5$

$4 a^{2} = 40$

$2 x^{2} = 24$

$x = \pm 2 \sqrt{3}$

For the following problems, solve for the indicated variable.

$x^{2} = 4 a^{2},$ for $x$

$x^{2} = 9 b^{2},$ for $x$

$x = \pm 3 b$

$a^{2} = 25 c^{2},$ for $a$

$k^{2} = m^{2} n^{2},$ for $k$

$k = \pm m n$

$k^{2} = p^{2} q^{2} r^{2},$ for $k$

$2 y^{2} = 2 a^{2} n^{2},$ for $y$

$y = \pm a n$

$9 y^{2} = 27 x^{2} z^{4},$ for $y$

$x^{2} - z^{2} = 0,$ for $x$

$x = \pm z$

$x^{2} - z^{2} = 0,$ for $z$

$5 a^{2} - 10 b^{2} = 0,$ for $a$

$a = b \sqrt{2}, - b \sqrt{2}$

For the following problems, solve each of the quadratic equations using the method of extraction of roots.

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

${(x - 2)}^{2} = 9$

$x = 5, - 1$

${(x - 3)}^{2} = 25$

${(a - 5)}^{2} = 36$

$x = 11, - 1$

${(a + 3)}^{2} = 49$

${(a + 9)}^{2} = 1$

$a = - 8, - 10$

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

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

$a = - 4 \pm \sqrt{5}$

${(b + 6)}^{2} = 7$

${(x + 1)}^{2} = a,$ for $x$

$x = - 1 \pm \sqrt{a}$

${(y + 5)}^{2} = b,$ for $y$

${(y + 2)}^{2} = a^{2},$ for $y$

$y = - 2 \pm a$

${(x + 10)}^{2} = c^{2},$ for $x$

${(x - a)}^{2} = b^{2},$ for $x$

$x = a \pm b$

${(x + c)}^{2} = a^{2},$ for $x$

calculator problems

For the following problems, round each result to the nearest hundredth.

$8 a^{2} - 168 = 0$

$a = \pm 4.58$

$6 m^{2} - 5 = 0$

$0.03 y^{2} = 1.6$

$y = \pm 7.30$

$0.048 x^{2} = 2.01$

$1.001 x^{2} - 0.999 = 0$

$x = \pm 1.00$

Exercises for review

( [link] ) Graph the linear inequality $3 (x + 2) < 2 (3 x + 4) .$

( [link] ) Solve the fractional equation $\frac{x - 1}{x + 4} = \frac{x + 3}{x - 1} .$

$x = \frac{- 11}{9}$

( [link] ) Find the product: $\sqrt{32 x^{3} y^{5}} \sqrt{2 x^{3} y^{3}} .$

( [link] ) Solve $x^{2} - 4 x = 0.$

$x = 0, 4$

( [link] ) Solve $y^{2} - 8 y = - 12.$

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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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Quadratic equations: solving quadratic equations using the method

Overview

The method of extraction of roots

Extraction of roots

The nature of solutions

Solutions of x 2 = K

Sample set a

Practice set a

Sample set b

Practice set b

Exercises

calculator problems

Exercises for review

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Solutions of $x^{2} = K$