Proofing: sqrt(7+sqrt(48))+(sqrt(7-sqrt(48))=4

[Dwellerofholes]

i need help proving that



(sqrt(7+sqrt(48)))+(sqrt(7-sqrt(48))) = 4

the limitations are that you cannot manipulate the right side...only the left..

so basically, i need help simplifying the left down to be 4

thanks..

 

[Fermat]

Square the lhs.

Hint:
(a+b)(a-b) = (a² - b²)

 

[Diane_]

Try multiplying the left by (sqrt(7 + sqrt(48))-(sqrt(7-sqrt(48)) on itself. Since that's 1, you won't be changing the value, but you'll find a few things simplifying.

 

[Dwellerofholes]

yea, i did that already, but the problem is, then i have complex radical denomintaors...

 

[arildno]

You should use Fermat's squaring technique.

 

[Dwellerofholes]

wait a sec, what is the lhc...to not change the value, wouldn't i have to multiply it over itself if i wanted to get the a^2 -B^2?

can you clarify what you mean?

 

[arildno]

Simplify the expression:
$$(\sqrt{7+\sqrt{48}}+\sqrt{7-\sqrt{48}})^{2}$$

 

[Dwellerofholes]

that would change the value...i would have to also square the 4 on the right side...which is against the rules of the problem

 

[arildno]

that would change the value...i would have to also square the 4 on the right side...which is against the rules of the problem

No.

Define "x" as follows:
$$x=\sqrt{7+\sqrt{48}}+\sqrt{7-\sqrt{48}}$$
Hence, we have:
$$x^{2}=(\sqrt{7+\sqrt{48}}+\sqrt{7-\sqrt{48}})^{2}$$
That is:
$$x^{2}=7+\sqrt{48}+2\sqrt{(\sqrt{7+\sqrt{48}})(\sqrt{7-\sqrt{48}})}+7-\sqrt{48}$$
that simplified reads:
$$x^{2}=14+2\sqrt{49-48}=14+2*1=16$$

 

[Dwellerofholes]

ok, thanks dude...i got as far as your third step before i realized that you posted again...i realized my error...

thanks for the great help guys!

 

[stmoe]

just do the whole squaring thing, and instead of having 4^2 put it all under a radical ... that way you get sqrt ( 16) = 4 .. which is true to some extent, because -4^2 = 16 as well ..

... i hate that little thing so much