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Radical equation: solve 2 * cbrt( 3x - 4 ) + 1 = 9
- Thread starter McCall
- Start date
You did it properly. There are several ways to check, but they are all basically the same and are as you did 'plug in the answer you got and see if it is true'. In this case, in order to avoid fractions, when you are checking, just stop at2 ³√3x-4 +1=9
2³3x-4=8
2 2
³√3x-4=4
(³√3x-4)³=(4)³
3x-4=64
3x=68
3x=68
3 3
x=22.67
how do you check this?
not sure if this is right, due to rounding.
2³√3(22.67)-4 +1 =9
2³√68.01-4 +1 =9
2³√64.01 +1=9
2(4)+1 =9
8+1=9
3x=68.
Then
2³√(68-4) +1 =9
2³√64 +1 =9
2³√64 +1=9
2(4)+1 =9
8+1=9
So
x=68/3=22.666...
How about
Instead of substituting x = 22.67 into the original equation substitute x = 68/3, which would check everything except your last division step. BTW if you get the wrong answer you can substitute x = 68/3 into any of the intermediary steps to see where you have gone wrong.
2 ³√3x-4 +1=9
2³3x-4=8
2 2
³√3x-4=4
(³√3x-4)³=(4)³
3x-4=64
3x=68
3x=68
3 3
x=22.67
how do you check this?
not sure if this is right, due to rounding.
2³√3(22.67)-4 +1 =9
2³√68.01-4 +1 =9
2³√64.01 +1=9
2(4)+1 =9
8+1=9
Instead of substituting x = 22.67 into the original equation substitute x = 68/3, which would check everything except your last division step. BTW if you get the wrong answer you can substitute x = 68/3 into any of the intermediary steps to see where you have gone wrong.
All of your work and methods seem fine and the checking process worked as intended. I just have two small points that might seem like I'm nitpicking, but as you go forward into harder math, noting these things will help you a lot. First, your equation needs some parentheses, to note which elements are under the radical. For instance, these two expressions produce totally different answers:
³√(68-4) = ³√(64) = 4
³√68 - 4 = ³√(68) - 4 ~ 0.0816...
In this case, I could easily tell from the context of your workings that the minus four was also meant to be under the radical, but it's not always clear. Secondly, I'd leave your answer in fraction form rather than rounding. 68/3 is really close to 22.67, but just enough off that some error crept in when you checked your answer. If you'd left it as 68/3, it would have worked out much nicer:
2*³√(3[68/3]-4)+1=9
2*³√(68-4)+1=9
...and so on
³√(68-4) = ³√(64) = 4
³√68 - 4 = ³√(68) - 4 ~ 0.0816...
In this case, I could easily tell from the context of your workings that the minus four was also meant to be under the radical, but it's not always clear. Secondly, I'd leave your answer in fraction form rather than rounding. 68/3 is really close to 22.67, but just enough off that some error crept in when you checked your answer. If you'd left it as 68/3, it would have worked out much nicer:
2*³√(3[68/3]-4)+1=9
2*³√(68-4)+1=9
...and so on
You did it properly. There are several ways to check, but they are all basically the same and are as you did 'plug in the answer you got and see if it is true'. In this case, in order to avoid fractions, when you are checking, just stop at
3x=68.
Then
2³√(68-4) +1 =9
2³√64 +1 =9
2³√64 +1=9
2(4)+1 =9
8+1=9
So
x=68/3=22.666...
Yes, that would be easier. Thanks for your help.
Instead of substituting x = 22.67 into the original equation substitute x = 68/3, which would check everything except your last division step. BTW if you get the wrong answer you can substitute x = 68/3 into any of the intermediary steps to see where you have gone wrong.
Good advise. Thanks for your help.