Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Mid-Chapter Check Point - Page 630: 19


No real solution; two imaginary solutions

Work Step by Step

Compute for the discriminant, $b^2-4ac$ $2x^2 + 5x + 4 = 0$ $a=2$, $b=5$, $c=4$ $b^2-4ac = 5^2-(4⋅2⋅4)$ $b^2-4ac = -7$ Note that the discriminant tells the number and types of solution. If: $b^2-4ac > 0$: Two unequal real solutions $b^2-4ac = 0$: One solution (a repeated solution) that is a real number $b^2-4ac < 0$: No real solution; two imaginary solutions The value of the discriminant of the given equation is $-7$, which is $<0$. Therefore, there are no real solution. When solved using the quadratic formula, two imaginary solutions will be obtained .
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