Question

misaka solved the radical equation ( x - 3=sqrt{4x - 7} ) but did not check her solutions.
misaka’s work
( (x - 3)^2=left(sqrt{4x - 7}
ight)^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} )?
options:
( x = 8 )
( x = 2 ) and ( x = 8 )
there are no true solutions to the equation.

Explanation:

Step1: Check x=2 in original equation

Substitute $x=2$ into $x-3=\sqrt{4x-7}$:
Left side: $2-3=-1$
Right side: $\sqrt{4(2)-7}=\sqrt{8-7}=\sqrt{1}=1$
$-1
eq 1$, so $x=2$ is extraneous.

Step2: Check x=8 in original equation

Substitute $x=8$ into $x-3=\sqrt{4x-7}$:
Left side: $8-3=5$
Right side: $\sqrt{4(8)-7}=\sqrt{32-7}=\sqrt{25}=5$
$5=5$, so $x=8$ is valid.

Answer:

$x=8$