Question

9-44. 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:

a. Equation: $(6x-18)(3x+2)=0$

Step1: Apply zero product rule

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

b. Equation: $x^2-7x+10=0$

Step1: Factor the quadratic

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

Step2: Apply zero product rule

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

Step3: Solve for $x$

$x=2$ or $x=5$

c. Equation: $2x^2+2x-12=0$

Step1: Simplify the equation

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

Step2: Factor the quadratic

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

Step3: Apply zero product rule

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

Step4: Solve for $x$

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

d. Equation: $4x^2-1=0$

Step1: Rewrite as difference of squares

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

Step2: Factor the expression

Use $a^2-b^2=(a-b)(a+b)$:
$(2x-1)(2x+1)=0$

Step3: Apply zero product rule

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

Step4: Solve for $x$

$x=\frac{1}{2}$ or $x=-\frac{1}{2}$

Answer:

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