Prove the expression

[Albi]

[MATH]\sqrt[4]{{79\pm 16\sqrt{15}} = \frac{\sqrt{30} +\sqrt{2}}{2}[/MATH]
Guys I'm having trouble to simplify the right side I don't know how to modify it to make it equal with the left side.

 

[pka]

[MATH]\sqrt[4]{79\pm 16\sqrt{15}} = \frac{\sqrt{30} +\sqrt{2}}{2}[/MATH]Guys I'm having trouble to simplify the right side I don't know how to modify it to make it equal with the left side.

I fixed your TeX coding. Is the question now correct?

 

[Albi]

I fixed your TeX coding. Is the question now correct?

Thank you very much.It is correct.
Could you tell me what did I do wrong in my code

 

[pka]

Could you tell me what did I do wrong in my code

You had a miss-match of {} on the \sqrt[4] function.
\(2^4\cdot(79\pm16\sqrt{15})=(\sqrt{32}+\sqrt{2})^4=2500\) See Here

 

[Albi]

You had a miss-match of {} on the \sqrt[4] function.
\(2^4\cdot(79\pm16\sqrt{15})=(\sqrt{32}+\sqrt{2})^4=2500\) See Here

Yeah we can deal with the problem now

 

[lookagain]

You had a miss-match of {} on the \sqrt[4] function.
\(2^4\cdot(79\pm16\sqrt{15})=(\sqrt{32}+\sqrt{2})^4=2500\) See Here

\(\displaystyle 2^4\cdot(79\pm16\sqrt{15}) \ \ne \ 2500\)

The expression on the left-hand side of the immediate line above is equal
to different values, but neither of them equals an integer. The largest
radicand is 30, not 32.

Albi's original equation in post # 1 is false because of the inclusion of the
plus or minus symbol.

It should be closer to this:

\(\displaystyle 2^4\cdot(79 + 16 \sqrt{15}) \ = \ (\sqrt{30} + \sqrt{2})^4 \ \approx \ 2255.48\)

 

[Albi]

\(\displaystyle 2^4\cdot(79\pm16\sqrt{15}) \ \ne \ 2500\)

The left-hand side of the immediate line above is equal to different values,
but neither of them equals an integer.

Albi's original equation in post # 1 is false.

The first post of pka where he corrected my latex code that is the problem that I'm trying to solve

 

[Dr.Peterson]

Yeah we can deal with the problem now

pka was proceeding toward the answer for you there; look closely!

Your equation really should be [MATH]\sqrt[4]{79\pm 16\sqrt{15}} = \frac{\sqrt{30}\pm \sqrt{2}}{2}[/MATH]
To prove it, multiply both sides by 2 and raise them to the fourth power. You will find that the equation is true (for both plus and minus cases).

To transform the RHS to the LHS, raise it to the fourth power and take the fourth root of the result (noting that the RHS is positive as it must be, in each case).

 

[lookagain]

The first post of pka where he corrected my latex code that is the problem that I'm trying to solve

It's not quite. If you include the plus or minus sign on the left-hand side, then you
must have it on the right- hand side in that expression.

Raise each side to the fourth power, or equivalently, square each side twice to show
that you can manipulate the right-hand side to be the radicand in the original left-hand side.

\(\displaystyle \bigg(\dfrac{\sqrt{30} \pm \sqrt{2}}{2}\bigg)^2 \ = \)

\(\displaystyle \dfrac{30 \pm 2\sqrt{60} + 2}{4} \ = \)

\(\displaystyle \dfrac{32 \pm 4\sqrt{15}}{4} \ =\)

\(\displaystyle 8 \pm \sqrt{15}\)

Now, you can square this result one more time to check that it matches the original
radicand with that fourth root.



(Dr.Peterson, I have been typing my response out one character at a time on a
tablet for many minutes, and I do not want to delete it even though its content
is similar to yours.)

 

[Albi]

Thanks everyone for helping me out

 

[Steven G]

I am confused once again. The lhs having the plus/minus will yield two values while the rhs will yield one number. Dr Peterson however put a plus/minus on the rhs. Where did it come from? Are you suggesting that what pka wrote was incorrect?

 

[lookagain]

I am confused once again. The lhs having the plus/minus will yield two values while the rhs will yield one number. Dr Peterson however put a plus/minus on the rhs. Where did it come from? Are you suggesting that what pka wrote was incorrect?

I think we should have asked Albi to go back and check what the exact problem
looked like.

 

[Albi]

I am confused once again. The lhs having the plus/minus will yield two values while the rhs will yield one number. Dr Peterson however put a plus/minus on the rhs. Where did it come from? Are you suggesting that what pka wrote was incorrect?

The expression is equal only for the pozitive value on the lhs pka wrote it correctly.