Certainly! Let's solve the equation [tex]\(\sqrt{2x + 1} = x - 1\)[/tex] step-by-step.
1. Isolate the square root on one side of the equation:
[tex]\(\sqrt{2x + 1} = x - 1\)[/tex]
2. Square both sides to eliminate the square root:
[tex]\(\left(\sqrt{2x + 1}\right)^2 = (x - 1)^2\)[/tex]
This simplifies to:
[tex]\(2x + 1 = (x - 1)^2\)[/tex]
3. Expand the right-hand side:
[tex]\((x - 1)^2 = x^2 - 2x + 1\)[/tex]
So the equation becomes:
[tex]\(2x + 1 = x^2 - 2x + 1\)[/tex]
4. Move all terms to one side to set the equation to zero:
[tex]\(0 = x^2 - 2x + 1 - 2x - 1\)[/tex]
Combine like terms:
[tex]\(0 = x^2 - 4x\)[/tex]
5. Factor the quadratic equation:
[tex]\(0 = x(x - 4)\)[/tex]
6. Set each factor equal to zero to find the solutions:
[tex]\(x = 0 \quad \text{or} \quad x = 4\)[/tex]
7. Verify each solution by substituting back into the original equation:
- For [tex]\(x = 0\)[/tex]:
[tex]\(\sqrt{2(0) + 1} = 0 - 1\)[/tex]
[tex]\(\sqrt{1} = -1\)[/tex]
[tex]\(1 \neq -1\)[/tex]
[tex]\(x = 0\)[/tex] is not a valid solution.
- For [tex]\(x = 4\)[/tex]:
[tex]\(\sqrt{2(4) + 1} = 4 - 1\)[/tex]
[tex]\(\sqrt{8 + 1} = 3\)[/tex]
[tex]\(\sqrt{9} = 3\)[/tex]
[tex]\(3 = 3\)[/tex]
[tex]\(x = 4\)[/tex] is a valid solution.
Thus, the solution to the equation [tex]\(\sqrt{2x + 1} = x - 1\)[/tex] is [tex]\(x = 4\)[/tex].
