Question

How do I solve:
(sqrt(2x^2-11x+14))^(1/3)=2-x

Answer 1

\[\sqrt{2x^2-11x+14}^\frac{1}{3}=2-x\]

Answer 2

Is that the problem?

Answer 3

yup!

Answer 4

So it is really:
\[(2x^2-11x+14)^{\frac{1}{6}}=2-x\]

Answer 5

is that 1/6?

Answer 6

yep

Answer 7

so would we multiple the 2-x by that too?

Answer 8

The only thing I can think of to do is to raise both sides to the 6th power.

Answer 9

ok

Answer 10

Proably use Pascal's triangle or the binomial theorem for the right side.

Answer 11

wait... if we multiplied the sq.root by 2 (so as to get 1/6), wouldn't we have to multiply the right by (1/6) as well?

Answer 12

wait that doesn't work... where did the 1/6 come from?

Answer 13

Would you agree that:
\[\sqrt{x}=x ^{\frac{1}{2}}\]

Answer 14

And would you further agree that you multiply exponents when raising to a power?

Answer 15

No, wouldn't it be \[\sqrt{x}=(x)^2\]

Answer 16

And so if you have:
\[(\sqrt{x})^{\frac{1}{3}}= [(x)^{\frac{1}{2}}]^{\frac{1}{3}}=x ^{\frac{1}{6}}\]

Answer 17

So you are saying that
\[\sqrt{x}=x^2\]

Answer 18

??????

Answer 19

So:
\[\sqrt{4}=4^2=16\] ???????

Answer 20

I always thought that the square root of 4 is 2 not 16

Answer 21

yes, if you want to get rid of a square root, you simply square it.

Answer 22

oh, you weren't trying to get RID of the square root.

Answer 23

So what you are really saying is that
\[(\sqrt{x})^2=x\]

Answer 24

So then yes, you're right.
yes, sorry.

Answer 25

That's a whole different thing than what you wrote before.

Answer 26

I guess it is! So could we not solve the question be first eliminating the square root?

Answer 27

Actually you have two roots: a square root and a third root

Answer 28

could you multiply the exponent of 1/3 by 3 (so as to cancel it) then just square the square root?