## College Algebra (10th Edition)

Published by Pearson

# Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding: 47

#### Answer

The solution set is {$1,\frac{3}{2}$}.

#### Work Step by Step

$2x^2-5x+3=0$ $a=2, b=-5, c=3$ $D=b^2-4ac=(-5)^2-4\cdot2(3)=25-24=1>0$ The equation has two real solutions. $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{5\pm\sqrt 1}{2\cdot2}$ Use the quadratic formula. $x=\frac{5\pm\sqrt 1}{4}$ Simplify $x=\frac{5+1}{4}=\frac{6}{4}=\frac{3}{2}$ $x=\frac{5-1}{4}=1$ The solution set is {$1,\frac{3}{2}$}.

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