Sure! Let's solve the equation [tex]\(\sqrt{3a - 5} = a - 3\)[/tex] step-by-step.
Step 1: Start with the given equation.
[tex]\[
\sqrt{3a - 5} = a - 3
\][/tex]
Step 2: To eliminate the square root, square both sides of the equation.
[tex]\[
(\sqrt{3a - 5})^2 = (a - 3)^2
\][/tex]
Simplify both sides:
[tex]\[
3a - 5 = a^2 - 6a + 9
\][/tex]
Step 3: Rearrange the equation to bring all terms to one side, setting it equal to zero.
[tex]\[
a^2 - 6a + 9 - 3a + 5 = 0
\][/tex]
Combine like terms:
[tex]\[
a^2 - 9a + 14 = 0
\][/tex]
Step 4: Solve the quadratic equation [tex]\(a^2 - 9a + 14 = 0\)[/tex]. To do this, we can factor the quadratic expression.
[tex]\[
a^2 - 9a + 14 = (a - 7)(a - 2) = 0
\][/tex]
Step 5: Set each factor equal to zero and solve for [tex]\(a\)[/tex].
[tex]\[
a - 7 = 0 \quad \text{or} \quad a - 2 = 0
\][/tex]
[tex]\[
a = 7 \quad \text{or} \quad a = 2
\][/tex]
Step 6: Check each solution in the original equation [tex]\(\sqrt{3a - 5} = a - 3\)[/tex] to verify if it holds true.
- For [tex]\(a = 7\)[/tex]:
[tex]\[
\sqrt{3(7) - 5} = 7 - 3
\][/tex]
[tex]\[
\sqrt{21 - 5} = 4
\][/tex]
[tex]\[
\sqrt{16} = 4
\][/tex]
[tex]\[
4 = 4 \quad \text{(True)}
\][/tex]
- For [tex]\(a = 2\)[/tex]:
[tex]\[
\sqrt{3(2) - 5} = 2 - 3
\][/tex]
[tex]\[
\sqrt{6 - 5} = -1
\][/tex]
[tex]\[
\sqrt{1} = -1
\][/tex]
[tex]\[
1 \neq -1 \quad \text{(False - this is not a valid solution)}
\][/tex]
Therefore, the only valid solution is [tex]\(a = 7\)[/tex].
Thus, the solution to the equation [tex]\(\sqrt{3a - 5} = a - 3\)[/tex] is:
[tex]\[
\boxed{7}
\][/tex]
