Question

algebra 2a semester online practice
complete this assessment to review what you’ve learned. it will not
solve the square root equation $x - 6 = \sqrt{x - 6}$. (1 point)
\\(\boldsymbol{x = 6}\\)
no solution
\\(\boldsymbol{x = 6}\\) and \\(\boldsymbol{x = 7}\\)
\\(\boldsymbol{x = 7}\\)

Explanation:

Step1: Let \( y = \sqrt{x - 6} \)

Then the equation \( x - 6 = \sqrt{x - 6} \) becomes \( y^2 = y \) (since \( x - 6 = (\sqrt{x - 6})^2 = y^2 \)).

Step2: Solve \( y^2 - y = 0 \)

Factor out \( y \): \( y(y - 1) = 0 \). So \( y = 0 \) or \( y = 1 \).

Step3: Substitute back \( y = \sqrt{x - 6} \)

• Case 1: \( \sqrt{x - 6} = 0 \)

Square both sides: \( x - 6 = 0 \), so \( x = 6 \).

• Case 2: \( \sqrt{x - 6} = 1 \)

Square both sides: \( x - 6 = 1 \), so \( x = 7 \).

Step4: Check solutions

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

Answer:

\( x = 6 \) and \( x = 7 \) (corresponding to the option " \( x = 6 \) and \( x = 7 \) ")