Question
The solutions to \( 15(2 a-2)=5\left(a^{2}-1\right) \) are:
Ask by Gonzalez George. in the United States
Mar 11,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The solutions are \( a=1 \) and \( a=5 \).
Solution
1. Expand the left-hand side:
\[
15(2a-2)=30a-30
\]
2. Expand the right-hand side:
\[
5\left(a^{2}-1\right)=5a^2-5
\]
3. Write the equation:
\[
30a-30=5a^2-5
\]
4. Rearrange the equation to set it equal to zero:
\[
0=5a^2-5-30a+30
\]
\[
0=5a^2-30a+25
\]
5. Divide the entire equation by 5:
\[
0=a^2-6a+5
\]
6. Factor the quadratic:
\[
a^2-6a+5=(a-1)(a-5)=0
\]
7. Set each factor equal to zero:
\[
a-1=0 \quad \text{or} \quad a-5=0
\]
8. Solve for \(a\):
\[
a=1 \quad \text{or} \quad a=5
\]
The solutions are \( a=1 \) and \( a=5 \).
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution Mind Expander
To solve \( 15(2 a-2)=5\left(a^{2}-1\right) \), we can start by simplifying the equation. Distributing both sides gives: \[ 30a - 30 = 5a^2 - 5. \] Next, rearranging all terms to one side yields: \[ 5a^2 - 30a + 25 = 0. \] Now, dividing the entire equation by 5 simplifies it to: \[ a^2 - 6a + 5 = 0. \] Factoring the quadratic: \[ (a - 5)(a - 1) = 0. \] Thus, the solutions are: \[ a = 5 \quad \text{and} \quad a = 1. \] So the solutions to the equation are \( a = 1 \) and \( a = 5 \).
