To determine if there is an extraneous solution to the equation [tex]\(\sqrt{x-3} = x - 5\)[/tex], let's carefully solve the equation step-by-step and check for any extraneous solutions.
1. Isolate the square root:
The given equation is [tex]\(\sqrt{x-3} = x - 5\)[/tex].
2. Square both sides:
To eliminate the square root, we square both sides of the equation:
[tex]\[
(\sqrt{x-3})^2 = (x - 5)^2
\][/tex]
Simplifying each side:
[tex]\[
x - 3 = (x - 5)^2
\][/tex]
3. Expand the squared term:
Expanding the right side of the equation:
[tex]\[
x - 3 = x^2 - 10x + 25
\][/tex]
4. Rearrange to form a quadratic equation:
Bring all terms to one side to set the equation to zero:
[tex]\[
0 = x^2 - 10x + 25 - (x - 3)
\][/tex]
Simplify:
[tex]\[
0 = x^2 - 11x + 28
\][/tex]
5. Solve the quadratic equation:
To solve [tex]\(x^2 - 11x + 28 = 0\)[/tex], we can use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -11\)[/tex], and [tex]\(c = 28\)[/tex].
The discriminant is:
[tex]\[
\Delta = (-11)^2 - 4 \cdot 1 \cdot 28 = 121 - 112 = 9
\][/tex]
Thus, the solutions are:
[tex]\[
x = \frac{11 \pm \sqrt{9}}{2} = \frac{11 \pm 3}{2}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{11 + 3}{2} = 7 \quad \text{and} \quad x = \frac{11 - 3}{2} = 4
\][/tex]
6. Check for extraneous solutions:
An extraneous solution is a solution that does not satisfy the original equation [tex]\(\sqrt{x-3} = x - 5\)[/tex] when substituted back.
- For [tex]\(x = 7\)[/tex]:
[tex]\[
\sqrt{7-3} = \sqrt{4} = 2 \quad \text{and} \quad 7 - 5 = 2
\][/tex]
Since both sides are equal, [tex]\(x = 7\)[/tex] is a valid solution.
- For [tex]\(x = 4\)[/tex]:
[tex]\[
\sqrt{4-3} = \sqrt{1} = 1 \quad \text{and} \quad 4 - 5 = -1
\][/tex]
Since both sides are not equal, [tex]\(x = 4\)[/tex] is an extraneous solution.
Thus, there is one valid solution [tex]\(x = 7\)[/tex]. The statement "Yes, the extraneous solution is [tex]\(x = 7\)[/tex]" is incorrect because [tex]\(x = 7\)[/tex] is actually a valid solution. The correct statement is:
- "Yes, the extraneous solution is [tex]\(x = 4\)[/tex]."
