Question

question 3
solve:
x(2x - 3)=0
ox = 3/2
ox = 0, x = 3/2
ox = 0, x = 2/3
ox = 1, x = 3/2
question 4
1 pts
solve by taking square roots:
(x + 1)^2 = 9
ox=-4,x = 2
ox=-4 only
o no solution
ox = 2 only

Explanation:

Question 3

Step1: Apply zero - product property

If $ab = 0$, then $a=0$ or $b = 0$. Here $a=x$ and $b = 2x-3$.

Step2: Solve for $x$ when $a = 0$

$x=0$

Step3: Solve for $x$ when $b = 0$

$2x-3=0$, then $2x=3$, so $x=\frac{3}{2}$

Question 4

Step1: Take square roots of both sides

$\sqrt{(x + 1)^2}=\pm\sqrt{9}$, so $x + 1=\pm3$

Step2: Solve for $x$ when $x + 1=3$

$x=3 - 1=2$

Step3: Solve for $x$ when $x + 1=-3$

$x=-3 - 1=-4$

Answer:

Question 3:

B. $x = 0, x=\frac{3}{2}$

Question 4:

A. $x=-4,x = 2$