Of course! Let's solve the equation [tex]\(\sqrt{x-1} + 3 = x\)[/tex] step by step.
1. Isolate the square root term:
[tex]\[
\sqrt{x - 1} + 3 = x
\][/tex]
Subtract 3 from both sides to isolate the square root term:
[tex]\[
\sqrt{x - 1} = x - 3
\][/tex]
2. Square both sides:
To eliminate the square root, we square both sides of the equation:
[tex]\[
(\sqrt{x - 1})^2 = (x - 3)^2
\][/tex]
This simplifies to:
[tex]\[
x - 1 = (x - 3)^2
\][/tex]
3. Expand the right-hand side:
Next, expand the right-hand side of the equation:
[tex]\[
x - 1 = (x - 3)(x - 3)
\][/tex]
[tex]\[
x - 1 = x^2 - 6x + 9
\][/tex]
4. Rearrange into a standard quadratic equation:
Bring all terms to one side to set the equation to zero:
[tex]\[
x - 1 - (x^2 - 6x + 9) = 0
\][/tex]
Simplify by combining like terms:
[tex]\[
x - 1 - x^2 + 6x - 9 = 0
\][/tex]
[tex]\[
-x^2 + 7x - 10 = 0
\][/tex]
Multiply through by -1 to make the quadratic term positive:
[tex]\[
x^2 - 7x + 10 = 0
\][/tex]
5. Factor the quadratic equation:
Factorize [tex]\(x^2 - 7x + 10\)[/tex]:
[tex]\[
(x - 2)(x - 5) = 0
\][/tex]
6. Solve for [tex]\(x\)[/tex]:
Set each factor equal to zero:
[tex]\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\][/tex]
[tex]\[
x - 5 = 0 \quad \Rightarrow \quad x = 5
\][/tex]
7. Check for extraneous solutions:
Substituting [tex]\(x = 2\)[/tex] back into the original equation:
[tex]\[
\sqrt{2 - 1} + 3 = 2
\][/tex]
[tex]\[
\sqrt{1} + 3 = 2
\][/tex]
[tex]\[
1 + 3 = 2 \quad \text{(false)}
\][/tex]
Therefore, [tex]\(x = 2\)[/tex] is an extraneous solution.
Substituting [tex]\(x = 5\)[/tex] back into the original equation:
[tex]\[
\sqrt{5 - 1} + 3 = 5
\][/tex]
[tex]\[
\sqrt{4} + 3 = 5
\][/tex]
[tex]\[
2 + 3 = 5 \quad \text{(true)}
\][/tex]
Therefore, [tex]\(x = 5\)[/tex] is a valid solution.
The solution to the equation [tex]\(\sqrt{x-1} + 3 = x\)[/tex] is:
[tex]\[
\boxed{5}
\][/tex]
