Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - 10.3 - Quadratic Formula - Concept Quiz 10.3 - Page 452: 5



Work Step by Step

The equation $x^{2}=27$ is actually $x^{2}+0x-27=0$. Since $x^{2}+0x-27=0$ is in the standard form of the quadratic equation $ax^{2}+bx+c=0$, it can be solved by the quadratic formula. We now solve this equation through the quadratic formula to illustrate our conclusion: Step 1: Comparing $x^{2}+0x-27=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=0$, and $c=-27$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a,b and c in the formula: $x=\frac{-(0) \pm \sqrt {(0)^{2}-4(1)(-27)}}{2(1)}$ Step 4: $x=\frac{0 \pm \sqrt {0+108}}{2}$ Step 5: $x=\frac{0 \pm \sqrt {108}}{2}$ Step 6: $x=\frac{\pm \sqrt {36\times3}}{2}$ Step 7: $x=\frac{\pm 6\sqrt 3}{2}$ Step 8: $x=\frac{-6\sqrt 3}{2}$ or $x=\frac{+6\sqrt 3}{2}$ Step 9: $x=-3\sqrt 3$ or $x=3\sqrt 3$ Step 10: Therefore, the solution set is {$-3\sqrt 3, 3\sqrt 3$}. Therefore, the question statement is true.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.