# Help with an proof

1. Oct 6, 2005

### 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..

2. Oct 6, 2005

### Fermat

Square the lhs.

Hint:
(a+b)(a-b) = (a² - b²)

3. Oct 6, 2005

### 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.

4. Oct 7, 2005

### Dwellerofholes

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

5. Oct 7, 2005

### arildno

You should use Fermat's squaring technique.

6. Oct 7, 2005

### Dwellerofholes

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

can you clarify what you mean?

7. Oct 7, 2005

### arildno

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

8. Oct 7, 2005

### 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

9. Oct 7, 2005

### arildno

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:
$$x^{2}=7+\sqrt{48}+2\sqrt{(\sqrt{7+\sqrt{48}})(\sqrt{7-\sqrt{48}})}+7-\sqrt{48}$$
$$x^{2}=14+2\sqrt{49-48}=14+2*1=16$$

10. Oct 7, 2005

### 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!!!!

11. Oct 12, 2005

### 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