Question 20 (2 points) Which of the following equations has both 3 and 6 as roots sqrt (x-2)=4-x b sqrt (x-2)=x-4 C sqrt (x-2)=(3)/(x) d sqrt (x-2)=(x)/(3)

To determine which equation has both 3 and 6 as roots, we need to substitute these values into each equation and see if they satisfy the equation.<br /><br />Let's start with option a:<br />$\sqrt {x-2}=4-x$<br /><br />Substituting x = 3:<br />$\sqrt {3-2}=4-3$<br />$\sqrt {1}=1$<br />This is true.<br /><br />Substituting x = 6:<br />$\sqrt {6-2}=4-6$<br />$\sqrt {4}=-2$<br />This is false.<br /><br />So, option a does not have both 3 and 6 as roots.<br /><br />Now let's move on to option b:<br />$\sqrt {x-2}=x-4$<br /><br />Substituting x = 3:<br />$\sqrt {3-2}=3-4$<br />$\sqrt {1}=-1$<br />This is false.<br /><br />Substituting x = 6:<br />$\sqrt {6-2}=6-4$<br />$\sqrt {4}=2$<br />This is true.<br /><br />So, option b does not have both 3 and 6 as roots.<br /><br />Next, let's check option c:<br />$\sqrt {x-2}=\frac {3}{x}$<br /><br />Substituting x = 3:<br />$\sqrt {3-2}=\frac {3}{3}$<br />$\sqrt {1}=1$<br />This is true.<br /><br />Substituting x = 6:<br />$\sqrt {6-2}=\frac {3}{6}$<br />$\sqrt {4}=\frac {1}{2}$<br />This is false.<br /><br />So, option c does not have both 3 and 6 as roots.<br /><br />Finally, let's check option d:<br />$\sqrt {x-2}=\frac {x}{3}$<br /><br />Substituting x = 3:<br />$\sqrt {3-2}=\frac {3}{3}$<br />$\sqrt {1}=1$<br />This is true.<br /><br />Substituting x = 6:<br />$\sqrt {6-2}=\frac {6}{3}$<br />$\sqrt {4}=2$<br />This is true.<br /><br />So, option d has both 3 and 6 as roots.<br /><br />Therefore, the correct answer is option d.