Hi. Im trying to figure out how to solve an equation for x. The equation is y=2(x-3)^2. If I were to work this out, I would have come up with the answer x=sqrt y/2 +3 but the answer is actually x=sqrt 2y/2 +3. I can't work out where the extra 2 has come from.
y = 2(x-3)^2
(y/2) = (x-3)^2
x-3 = sqrt(y/2)
x = sqrt(y/2) + 3 is the answer.
Was the "square root" shown as a radical sign with a 2 in front? I makes no sense to write sqrt(2y/2). That would be just sqrt(y)
To solve the equation y=2(x-3)^2 for x, let's go step by step:
1. Start by isolating the squared term (x-3)^2 by dividing both sides of the equation by 2:
y/2 = (x-3)^2
2. Take the square root of both sides to eliminate the exponent of 2. Remember to include both the positive and negative square roots:
sqrt(y/2) = ±sqrt((x-3)^2)
3. Now let's simplify the right side of the equation by applying the square root to (x-3)^2:
sqrt(y/2) = ±(x-3)
4. Move the -3 over to the left side by adding 3 to both sides of the equation:
sqrt(y/2) + 3 = ±x
5. Finally, since x is the variable we want to solve for, we can rewrite the equation as:
x = ±sqrt(y/2) + 3
Thus, the correct answer is x = ±sqrt(y/2) + 3. The extra "2" in the square root isn't necessary based on the given equation y=2(x-3)^2. Double-check your understanding of the problem and make sure you've performed the steps correctly.