Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.6 Solving Quadratic Equations by Using the Quadratic Formula - 4.6 Exercises - Page 373: 67

Answer

$x=1.5-0.5\sqrt{105}, x=1.5+0.5\sqrt{105}$

Work Step by Step

Use the completing the square method. $$ \begin{aligned} \frac{1}{4} x^2-\frac{3}{4} x+7 & =13 \\ \frac{x^2}{4}-\frac{3 x}{4} & =13-7 \\ \left(\frac{x^2}{4}-\frac{3 x}{4}\right) \cdot 4 & =6(4)\\ x^2-3 x+1.5^2 & =24+1.5^2 \\ (x-1.5)^2 & =26.25 \\ x-1.5 & = \pm \sqrt{26.25} \\ x & = 1.5 \pm 0.5\sqrt{105}. \end{aligned} $$ This gives the solution: $$ \begin{aligned} x & =1.5-0.5\sqrt{105} \\ & \approx -3.62347 \\ x & =1.5+0.5\sqrt{105} \\ & \approx 6.62347. \end{aligned} $$ Define the following version of the function and plot it. We see that the zeros are where they should be. $$ f(x) =(x-1.5)^2-26.25. $$
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.