Question
Solve the equation. Express numbers in exact simplest form. \[ 3 y(y+3)=12 \] The solution set is \( \{\square\} \).
Ask by Savage Davies. in the United States
Mar 09,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The solutions are \( y = -4 \) and \( y = 1 \).
Solution
We start with the equation:
\[
3y(y+3) = 12
\]
**Step 1. Expand the left-hand side**
\[
3y^2 + 9y = 12
\]
**Step 2. Bring all terms to one side**
\[
3y^2 + 9y - 12 = 0
\]
**Step 3. Divide the entire equation by 3 to simplify**
\[
y^2 + 3y - 4 = 0
\]
**Step 4. Factor the quadratic equation**
We look for two numbers that multiply to \(-4\) and add to \(3\). These numbers are \(4\) and \(-1\).
\[
y^2 + 3y - 4 = (y + 4)(y - 1)
\]
**Step 5. Use the Zero-Product Property**
Set each factor equal to zero:
\[
y + 4 = 0 \quad \text{or} \quad y - 1 = 0
\]
Thus, the solutions are:
\[
y = -4 \quad \text{or} \quad y = 1
\]
**Answer:**
The solution set is \( \{-4, 1\} \).
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Simplify this solution The Deep Dive
To solve the equation \( 3y(y+3) = 12 \), we first simplify it: 1. Divide both sides by 3: \[ y(y+3) = 4. \] 2. Expand the left side: \[ y^2 + 3y - 4 = 0. \] 3. Now, we can factor the quadratic: \[ (y + 4)(y - 1) = 0. \] 4. Set each factor to zero: \[ y + 4 = 0 \quad \Rightarrow \quad y = -4, \] \[ y - 1 = 0 \quad \Rightarrow \quad y = 1. \] Thus, the solution set is \(\{-4, 1\}\).
