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.Let's start with option a: Substituting x = 3: This is true.Substituting x = 6: This is false.So, option a does not have both 3 and 6 as roots.Now let's move on to option b: Substituting x = 3: This is false.Substituting x = 6: This is true.So, option b does not have both 3 and 6 as roots.Next, let's check option c: Substituting x = 3: This is true.Substituting x = 6: This is false.So, option c does not have both 3 and 6 as roots.Finally, let's check option d: Substituting x = 3: This is true.Substituting x = 6: This is true.So, option d has both 3 and 6 as roots.Therefore, the correct answer is option d.