Question

misaka solved the radical equation $x - 3 = \sqrt{4x - 7}$ but did not check her solutions.\\
$(x - 3)^2 = (\sqrt{4x - 7})^2$\\
$x^2 - 6x + 9 = 4x - 7$\\
$x^2 - 10x + 16 = 0$\\
$(x - 2)(x - 8) = 0$\\
$x = 2$ and $x = 8$\\
which shows the true solution(s) to the radical equation $x - 3 = \sqrt{4x - 7}$?\\
\bigcirc $x = 2$\\
\bigcirc $x = 8$\\
\bigcirc $x = 2$ and $x = 8$\\
\bigcirc there are no true solutions to this equation.

Explanation:

Step 1: Check \( x = 2 \)

Substitute \( x = 2 \) into the original equation \( x - 3=\sqrt{4x - 7} \).
Left - hand side (LHS): \( 2-3=-1 \)
Right - hand side (RHS): \( \sqrt{4\times2 - 7}=\sqrt{8 - 7}=\sqrt{1}=1 \)
Since \( - 1
eq1 \), \( x = 2 \) is not a solution.

Step 2: Check \( x = 8 \)

Substitute \( x = 8 \) into the original equation \( x - 3=\sqrt{4x - 7} \).
Left - hand side (LHS): \( 8 - 3=5 \)
Right - hand side (RHS): \( \sqrt{4\times8 - 7}=\sqrt{32 - 7}=\sqrt{25}=5 \)
Since \( 5 = 5 \), \( x = 8 \) is a solution.

Answer:

B. \( x = 8 \)