Math  /  Algebra

QuestionSolve for xx. Enter the solutions from least to greatest. (2x+4)(3x2)=0(2 x+4)(3 x-2)=0 lesser x=x= \square greater x=x= \square

Studdy Solution

STEP 1

What is this asking? We're looking for the values of xx that make this equation (2x+4)(3x2)=0(2x+4)(3x-2)=0 true. Watch out! Remember, if two things multiplied together equal zero, then at least one of those things *must* be zero!

STEP 2

1. Set each factor to zero.
2. Solve for xx in each equation.

STEP 3

Alright, so we've got this equation (2x+4)(3x2)=0(2x+4)(3x-2)=0.
Since the product is **zero**, either 2x+42x+4 must be zero, *or* 3x23x-2 must be zero.
Maybe both!
Let's set each factor equal to zero and solve separately.
This gives us two new equations: 2x+4=02x+4=0 and 3x2=03x-2=0.

STEP 4

Let's tackle 2x+4=02x+4=0 first.
We want to isolate xx, so let's subtract 4 from both sides of the equation to keep it balanced.
This gives us 2x+44=042x + 4 - 4 = 0 - 4, which simplifies to 2x=42x = -4.
Now, we divide both sides by **2** (because we're multiplying xx by 2, so to get xx by itself we divide to one) to get 2x2=42\frac{2x}{2} = \frac{-4}{2}.
This simplifies to x=2x = -2.
Boom! Our first solution is x=2x = \mathbf{-2}.

STEP 5

Now, let's conquer 3x2=03x-2=0.
We'll add 2 to both sides to get 3x2+2=0+23x - 2 + 2 = 0 + 2, which simplifies to 3x=23x=2.
Dividing both sides by **3** gives us 3x3=23\frac{3x}{3} = \frac{2}{3}, so x=23x = \frac{2}{3}.
Awesome! Our second solution is x=23x = \mathbf{\frac{2}{3}}.

STEP 6

We found two solutions: x=2x = -2 and x=23x = \frac{2}{3}.
Since the problem asks for the solutions from least to greatest, our final answer is: lesser x=2x = \mathbf{-2} and greater x=23x = \mathbf{\frac{2}{3}}.

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