night help: How do you solve this? - Help.com

Is that the whole problem? It doesn’t look complete.

Yeah, that’s the whole problem. The lesson it’s in is for solving radical equations if that helps. :/

If you cube both sides of the equation, the answer should come naturally.

I tried but didn’t get the right answer. I dont know what im doing wrong :(

it’s multiple choice.. the answers are either

a) x=51
b) x=9
c) x=79
d) x=-89

and i got none of them :/

Sorry, I did’t get any of those either, I got a really big number :(

It’s possible you copied it incorrectly and
^3√x-2^+7=28 should be: 3(x-2)+7=28?
In this case, you would get one of those answers.

DarkSnow wrote:
It’s possible you copied it incorrectly and
^3√x-2^+7=28 should be: 3(x-2)+7=28?
In this case, you would get one of those answers.

Or…
It could have been 3*sqrt(x-2)+7=28, in which case you would get another one of those answers.

DarkSnow wrote:

DarkSnow wrote:
It’s possible you copied it incorrectly and
^3√x-2^+7=28 should be: 3(x-2)+7=28?
In this case, you would get one of those answers.

Or…
It could have been 3*sqrt(x-2)+7=28, in which case you would get another one of those answers.

I’m curious how to solve this one. I think the square root is throwing me off.

If you knew what X was, the order of mathematical operations is PEMDAS:
1) Parentheses (the x-2 term)
2) Exponents (square root is the power of 0.5)
3) Multiplication (the “3*” part)
4) Division (none here)
5) Addition (+7)
6) Subtraction (none here)

However, in this case we don’t know what X is, we want to find it out. So we have to work backwards and cancel out the operations that would have been done otherwise. So you undo any subtraction (but there is no subtraction). Then you undo the addition of +7 by subtracting by 7. However, you have to change both sides by the same amount so that the equals sign remains true:
3*sqrt(x-2)+7-7 = 28-7 –> 3*sqrt(x-2) = 21.

Similarly you do the other steps in reverse. So it would go like this:
6) None
5) 3*sqrt(x-2)+7-7=28-7 –> 3*sqrt(x-2)=21
4) None
3) 3/3*sqrt(x-2)=21/3 –> sqrt(x-2)=7
2) sqrt(x-2) is the same as (x-2)^0.5, so [(x-2)^0.5]^2=7^2 –> (x-2)^1 = 49 –> x-2=49
1) Evaluate the parentheses. You go through the order of steps again, but in this case you do 6) and then skip 5) through 1) since you have to undo a subtraction: x-2,
So x-2+2=49+2 –> x=51

Going through this process in an organized and logical fashion is time-consuming. But with enough practice people can do this in their head in a few seconds. ^^

I think you’re right, powers are the oddest part of these problems, especially square root and other fraction powers. There are just a few rules for them that you have to memorize (or you can remember how to determine them on your own). When you have something such as (2^2)^2 it’s the same as 2^(2*2) = 2^4, because raising something to a power and then the result of that to a power again multiplies out.
Square root is the power of 1/2 or 0.5, so sqrt(2) = 2^0.5. Similarly, the cube root is the power of 1/3. So to undo a square root in the equation you have to raise both sides by the opposite power, 2 in this case. Or, for example, if it was (x-2)^(1/8) you’d have to raise both sides to the power of 8 to undo it. Because [(x-2)^0.5]^2 = (x-2)^(0.5*2) = (x-2)^(1) = (x-2) = x-2. And of course you have to do the same thing to the other side when you do this, and you have 7^2 = 49.

Thanks, DarkSnow. I think I understand it now. Once I got to the square root step, I couldn’t remember what needed to be done. Now I remember that the opposite of the square root must be performed for both sides.

Right. No problem. ^^