College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Review Exercises: 52

Answer

$\text{Discriminant: } 484 \\\text{Number of distinct solutions: } 2 \\\text{Type of solutions: rational solutions }$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To evaluate the discriminant of the given equation, $ 16x^2+3=-26x ,$ identify first the values of $a,b,$ and $c$ in the quadratic expression $ax^2+bx+c.$ Then use the Discriminant Formula. If the value of the discriminant is less than zero, then there are $\text{ 2 nonreal complex solutions .}$ If the value is $0,$ then there is $\text{ 1 distinct rational solution .}$ If the value of the discriminant is a positive perfect square, then there are $\text{ 2 rational solutions .}$ Finally, if the value of the discriminant is positive but not a perfect square, there are $\text{ 2 irrational solutions .}$ $\bf{\text{Solution Details:}}$ In the form $ax^2+bx+c=0,$ the given equation is equivalent to \begin{array}{l}\require{cancel} 16x^2+26x+3=0 .\end{array} In the equation above, $a= 16 ,$ $b= 26 ,$ and $c= 3 .$ Using the Discriminant Formula which is given by $b^2-4ac,$ the value of the discriminant is \begin{array}{l}\require{cancel} (26)^2-4(16)(3) \\\\= 676-192 \\\\= 484 \\\\= 22^2 .\end{array} Since the discriminant is $\text{ is greater than zero and a perfect square ,}$ then there are $\text{ 2 rational solutions .}$ Hence, the given equation has the following properties: \begin{array}{l}\require{cancel} \text{Discriminant: } 484 \\\text{Number of distinct solutions: } 2 \\\text{Type of solutions: rational solutions } .\end{array}
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