Question

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

Explanation:

Step1: Let \( y = \sqrt{x - 6} \), then the equation becomes \( y = y^2 \) (since \( \sqrt{x - 6}=y \) and \( x - 6 = y^2 \) from the square root definition, so substituting back we get \( y = y^2 \))

\( y^2 - y = 0 \)

Step2: Factor the equation

\( y(y - 1) = 0 \)

Step3: Solve for \( y \)

So \( y = 0 \) or \( y - 1 = 0 \) which gives \( y = 0 \) or \( y = 1 \)

Step4: Substitute back \( y = \sqrt{x - 6} \)

• Case 1: When \( y = 0 \), \( \sqrt{x - 6}=0 \)

Squaring both sides: \( x - 6 = 0 \) so \( x = 6 \)

• Case 2: When \( y = 1 \), \( \sqrt{x - 6}=1 \)

Squaring both sides: \( x - 6 = 1 \) so \( x = 7 \)

Step5: Check the solutions in the original equation

• For \( x = 6 \): Left side \( \sqrt{6 - 6}=\sqrt{0}=0 \), Right side \( 6 - 6 = 0 \). So \( x = 6 \) is valid.
• For \( x = 7 \): Left side \( \sqrt{7 - 6}=\sqrt{1}=1 \), Right side \( 7 - 6 = 1 \). So \( x = 7 \) is valid.

Answer:

\( x = 6 \) or \( x = 7 \)