Question

5-44. solve the equations. check your solutions. homework help
a. $(6x - 18)(3x + 2) = 0$
b. $x^2 - 7x + 10 = 0$
c. $2x^2 + 2x - 12 = 0$
d. $4x^2 - 1 = 0$

Explanation:

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Part a

Step1: Apply zero product property

Set each factor equal to 0:
$6x - 18 = 0$ or $3x + 2 = 0$

Step2: Solve first linear equation

Isolate $x$:
$6x = 18 \implies x = \frac{18}{6} = 3$

Step3: Solve second linear equation

Isolate $x$:
$3x = -2 \implies x = -\frac{2}{3}$

Step4: Verify solutions

Substitute $x=3$: $(6*3-18)(3*3+2)=(18-18)(11)=0*11=0$
Substitute $x=-\frac{2}{3}$: $(6*(-\frac{2}{3})-18)(3*(-\frac{2}{3})+2)=(-4-18)(-2+2)=(-22)(0)=0$

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Part b

Step1: Factor the quadratic

Find two terms that multiply to 10 and add to -7:
$x^2 -7x +10=(x-2)(x-5)=0$

Step2: Apply zero product property

Set each factor equal to 0:
$x-2=0$ or $x-5=0$

Step3: Solve for $x$

$x=2$ or $x=5$

Step4: Verify solutions

Substitute $x=2$: $2^2 -7*2 +10=4-14+10=0$
Substitute $x=5$: $5^2 -7*5 +10=25-35+10=0$

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Part c

Step1: Simplify the quadratic

Divide all terms by 2:
$x^2 +x -6=0$

Step2: Factor the simplified quadratic

Find two terms that multiply to -6 and add to 1:
$(x+3)(x-2)=0$

Step3: Apply zero product property

Set each factor equal to 0:
$x+3=0$ or $x-2=0$

Step4: Solve for $x$

$x=-3$ or $x=2$

Step5: Verify solutions

Substitute $x=-3$: $2*(-3)^2 +2*(-3)-12=18-6-12=0$
Substitute $x=2$: $2*(2)^2 +2*(2)-12=8+4-12=0$

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Part d

Step1: Rewrite as difference of squares

Express in $a^2 - b^2$ form:
$4x^2 -1=(2x)^2 -(1)^2=(2x-1)(2x+1)=0$

Step2: Apply zero product property

Set each factor equal to 0:
$2x-1=0$ or $2x+1=0$

Step3: Solve for $x$

$2x=1 \implies x=\frac{1}{2}$; $2x=-1 \implies x=-\frac{1}{2}$

Step4: Verify solutions

Substitute $x=\frac{1}{2}$: $4*(\frac{1}{2})^2 -1=4*\frac{1}{4}-1=1-1=0$
Substitute $x=-\frac{1}{2}$: $4*(-\frac{1}{2})^2 -1=4*\frac{1}{4}-1=1-1=0$

Answer:

a. $x=3$ and $x=-\frac{2}{3}$
b. $x=2$ and $x=5$
c. $x=-3$ and $x=2$
d. $x=\frac{1}{2}$ and $x=-\frac{1}{2}$