College Algebra (10th Edition)

The solution set is $\left\{\frac{1}{2}, \frac{3}{2}\right\}$.
Distribute $2$ to obtain: $4u^2-8u+3=0$ 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 $a=4$, $b=-8$, and $c=3$. Thus, $ac = 4(3) = 12$ Note that $12=-6(-2)$ and $-8 = (-6)+(-2)$ This means that $d=-6$ and $e=-2$. Rewrite the middle term of the trinomial as $-6u$ +$(-2u)$ to obtain: $4u^2-8u+3 = 0 \\4u^2+(-6u)+(-2u)+3 = 0 \\4u^2-6u+(-2u) +3=0$ Regroup then factor out the GCF in each group to obtain: $(4u^2-6u)+(-2u+3)=0 \\2u(2u-3) +(-1)(2u-3)=0$ Factor out $2u-3$ to obtain: $(2u-3)(2u-1)=0$ Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain: $\begin{array}{ccc} &2u-3=0 &\text{ or } &2u-1=0 \\&2u=3 &\text{ or } &2u=1 \\&u=\frac{3}{2} &\text{ or } &u=\frac{1}{2} \end{array}$ Therefore the solution set is $\left\{\frac{1}{2}, \frac{3}{2}\right\}$.