Question

01/20/26 unit 8 review day 1 - solving and solve the equation. check for extraneous solutions.
\\(\sqrt{x} + 6 = x\\)
\\(\circ\\) 16
\\(\circ\\) 4, 9
\\(\circ\\) 4
\\(\circ\\) 25
\\(\circ\\) 9

Explanation:

Step1: Isolate the square root

Subtract 6 from both sides of the equation $\sqrt{x}+6 = x$ to get $\sqrt{x}=x - 6$.

Step2: Square both sides

Square both sides of the equation $\sqrt{x}=x - 6$ to eliminate the square root. We have $(\sqrt{x})^2=(x - 6)^2$, which simplifies to $x=x^{2}-12x + 36$.

Step3: Rearrange into quadratic equation

Rearrange the equation $x=x^{2}-12x + 36$ to standard quadratic form $x^{2}-13x + 36 = 0$.

Step4: Factor the quadratic equation

Factor the quadratic equation $x^{2}-13x + 36 = 0$. We need two numbers that multiply to 36 and add up to - 13. The numbers are -4 and -9. So, $(x - 4)(x - 9)=0$.

Step5: Solve for x

Set each factor equal to zero: $x - 4=0$ gives $x = 4$; $x - 9=0$ gives $x = 9$.

Step6: Check for extraneous solutions

- For $x = 4$: Substitute into the original equation $\sqrt{4}+6=2 + 6=8
eq4$. So, $x = 4$ is extraneous.
- For $x = 9$: Substitute into the original equation $\sqrt{9}+6=3 + 6=9$. So, $x = 9$ is a valid solution. Wait, but wait, let's check again. Wait, the options have 9 as an option? Wait, no, wait, maybe I made a mistake. Wait, let's re - check the squaring step.

Wait, when we squared $\sqrt{x}=x - 6$, we assumed that $x-6\geq0$ (since the square root is non - negative). So $x\geq6$.

For $x = 4$: $x-6=4 - 6=-2<0$, and $\sqrt{4}=2\geq0$, but $x - 6$ is negative, so the equation $\sqrt{x}=x - 6$ is not valid for $x = 4$ because the left - hand side $\sqrt{x}$ is non - negative and the right - hand side $x - 6$ is negative when $x = 4$.

For $x=9$: $x - 6=9 - 6 = 3\geq0$, and $\sqrt{9}=3$, so $3=3$, which is valid. But wait, the options have 9? Wait, the options are 16, 4,9, 4,9, 4,25,9. Wait, maybe I made a mistake in the initial equation. Wait, the original equation is $\sqrt{x}+6=x$. Wait, let's re - solve:

Wait, step 1: $\sqrt{x}=x - 6$. Then $x-6\geq0\Rightarrow x\geq6$.

When we squared, we got $x=(x - 6)^2=x^{2}-12x + 36\Rightarrow x^{2}-13x + 36 = 0\Rightarrow(x - 4)(x - 9)=0$.

For $x = 9$: $\sqrt{9}+6=3 + 6=9$, which works.

For $x = 4$: $\sqrt{4}+6=2 + 6=8
eq4$, so it's extraneous. But wait, the options have 9 as an option? Wait, maybe the user made a typo in options, but according to the solution, $x = 9$ is the solution. But wait, let's check the options again. The options are: 16, 4,9, 4,9, 4,25,9. Wait, maybe I misread the equation. Wait, is the equation $\sqrt{x + 6}=x$? No, the equation is $\sqrt{x}+6=x$. Wait, if the equation was $\sqrt{x + 6}=x$, then:

Step 1: Square both sides: $x + 6=x^{2}\Rightarrow x^{2}-x - 6=0\Rightarrow(x - 3)(x + 2)=0\Rightarrow x = 3$ or $x=-2$. Then check: for $x = 3$, $\sqrt{3 + 6}=\sqrt{9}=3$, valid; for $x=-2$, $\sqrt{-2 + 6}=\sqrt{4}=2
eq - 2$, extraneous. But that's not the case here.

Wait, going back, the original equation is $\sqrt{x}+6=x$. We found $x = 9$ is valid. But in the options, 9 is present. Wait, maybe I made a mistake in the extraneous check for $x = 9$. Wait, $\sqrt{9}+6=3 + 6=9$, which is equal to $x = 9$. So $x = 9$ is valid.

Wait, but let's check the options again. The options are: 16, 4,9, 4,9, 4,25,9. So the correct answer should be 9? Wait, no, wait, maybe I made a mistake in the factoring. Wait, $x^{2}-13x + 36$: $13x$: 4 and 9, 4 + 9 = 13. So $(x - 4)(x - 9)=x^{2}-13x + 36$, that's correct.

Wait, when $x = 9$, $\sqrt{9}+6=3 + 6=9$, which is equal to $x$. So $x = 9$ is a solution. When $x = 4$, $\sqrt{4}+6=2 + 6=8
eq4$, so it's extraneous. So the solution is $x = 9$. But wait, the options have 9 as an option.

Wait, maybe the initial equation was $\sqrt{x + 6}=x$? Let's check that. If the equation is $\sqrt{x + 6}=x$:

Step 1: Square both sides: $x + 6=x^{2}\Rightarrow x^{2}-x - 6=0\Rightarrow(x - 3)(x + 2)=0\Rightarrow x = 3$ or $x=-2$.

Check $x = 3$: $\sqrt{3 + 6}=\sqrt{9}=3$, valid.

Check $x=-2$: $\sqrt{-2 + 6}=\sqrt{4}=2
eq - 2$, extraneous. But 3 is not in the options. So the original equation must be $\sqrt{x}+6=x$. Then the solution is $x = 9$.

Answer:

9