College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 101: 23


The solution set is $\left\{\frac{3}{2}\right\}$.

Work Step by Step

Subtract $12x$ to both sides of the equation to obtain: $4x^2-12x+9=0$ RECALL: A quadratic trinomial $ax^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = ac$ and $d+e =b$ If such integers exist, then $ax^2+bx + c = ax^2+dx + ex + c$, and may be factored by grouping. The trinomial in the equation has $a=4$, $b=-12$, and $c=9$. Thus, $ac = 4(9) = 36$ Note that $36=-6(-6)$ and $-12 = (-6)+(-6)$ This means that $d=-6$ and $e=-6$. Rewrite the middle term of the trinomial as $-6x$ +$(-6x)$ to obtain: $4x^2-12x+9 = 0 \\4x^2+(-6x)+(-6x)+9 = 0 \\4x^2-6x+(-6x) +9=0$ Regroup then factor out the GCF in each group to obtain: $(4x^2-6x)+(-6x+9)=0 \\2x(2x-3) +(-3)(2x-3)=0$ Factor out $2x-3$ to obtain: $(2x-3)(2x-3)=0$ Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain: $\begin{array}{ccc} &2x-3=0 &\text{ or } &2x-3=0 \\&2x=3 &\text{ or } &2x=3 \\&x=\frac{3}{2} &\text{ or } &x=\frac{3}{2} \end{array}$ Therefore the solution set is $\left\{\frac{3}{2}\right\}$.
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