Question

question
solve for all possible values of x.
\\(\sqrt{6 - x} = x - 6\\)
answer attempt 1 out of 2
\\(\oplus\\) additional solution \\(\circleddash\\) no solution
\\(x = \square\\) submit answer

Explanation:

Step1: Analyze the domain of the square root

For the square root $\sqrt{6 - x}$ to be defined, the expression inside the square root must be non - negative. So we have the inequality $6 - x\geq0$, which implies $x\leq6$. Also, the right - hand side of the equation is $x - 6$. Let's consider the non - negativity of the square root: $\sqrt{6 - x}\geq0$. So the right - hand side $x - 6$ must also be non - negative (because the left - hand side is non - negative and they are equal), so $x-6\geq0$, which implies $x\geq6$.

Step2: Find the intersection of the two domain conditions

From $x\leq6$ (from the square root definition) and $x\geq6$ (from the non - negativity of the right - hand side), the only value that satisfies both is $x = 6$.

Step3: Verify the solution

Substitute $x = 6$ into the original equation:
Left - hand side: $\sqrt{6 - 6}=\sqrt{0}=0$
Right - hand side: $6 - 6 = 0$
Since the left - hand side equals the right - hand side when $x = 6$, $x = 6$ is a solution. But we also need to check if there are other solutions. Let's square both sides of the original equation $\sqrt{6 - x}=x - 6$ to eliminate the square root.
Squaring both sides gives $6 - x=(x - 6)^2$. Expand the right - hand side: $(x - 6)^2=x^{2}-12x + 36$. So the equation becomes $6 - x=x^{2}-12x + 36$.
Rearrange this equation to get a quadratic equation: $x^{2}-12x + 36-(6 - x)=0$, which simplifies to $x^{2}-11x + 30 = 0$.
Factor the quadratic equation: $x^{2}-11x + 30=(x - 5)(x - 6)=0$. The solutions of the quadratic equation are $x = 5$ and $x = 6$. But we already know from the domain analysis that $x$ must satisfy $x\leq6$ and $x\geq6$, so $x = 5$ does not satisfy $x\geq6$ (because when $x = 5$, the right - hand side $x - 6=5 - 6=-1$ and the left - hand side $\sqrt{6 - 5}=1$, and $1
eq - 1$). So we discard $x = 5$.

Answer:

$x = 6$