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) x=2 x=8 x=2 and x=8 There are no true solutions to the equation.

To find the true solution(s) to the radical equation $x-3=\sqrt {4x-7}$, we need to check the solutions obtained from squaring both sides of the equation.<br /><br />The solutions obtained are $x=2$ and $x=8$. However, we need to verify if these solutions satisfy the original equation.<br /><br />Let's substitute $x=2$ into the original equation:<br />$2-3=\sqrt {4(2)-7}$<br />$-1=\sqrt {1}$<br />$-1=1$<br /><br />This is not true, so $x=2$ is not a valid solution.<br /><br />Now, let's substitute $x=8$ into the original equation:<br />$8-3=\sqrt {4(8)-7}$<br />$5=\sqrt {25}$<br />$5=5$<br /><br />This is true, so $x=8$ is a valid solution.<br /><br />Therefore, the true solution to the radical equation $x-3=\sqrt {4x-7}$ is $x=8$.