## Callisto Group Title Tutor question #3 Simplify: $\sqrt{5-\sqrt{21}}$ (conditions: pen, paper, no calculator, done in 2 minutes) * 2 years ago 2 years ago

1. lgbasallote

what if x = $$\sqrt{5-\sqrt{21}}$$ then $$x^2 = 5 - \sqrt{21}$$ $x^2 - 5 = -\sqrt{21}$ $(x^2 - 5)^2 = 21$ oh goodness quadratic o.O

2. satellite73

actually it is a 4th degree equation

3. Callisto

$(x^2-5)^2=21$$x^4-10x^2+25=21$$x^4-10x^2+4=0$$x^4-10x^2+4=0$$x^2=\frac{10\pm \sqrt{(-10)^6-4(4)}}{2}$$x^2=5\pm \sqrt{21}$Sorry... are you sure that it is 'simplified'?

4. Callisto

Oh.that's power 2, typo

5. satellite73

apparently it is also $\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}$ i remember seeing something like this before but i am not sure i remember how to go from one to the other

6. Callisto

I need to know how to work that out.....

7. satellite73

cant seem to get it, i think it was a trick

8. Callisto

............................ It... was ... a ... question ... asked ... when ... my friend applied for a summer job ...

9. satellite73

really? doing what??

10. Callisto

tutor - teaching high school students

11. satellite73

hmm i wonder what answer they wanted

12. Callisto

Same here...

13. Callisto

When I was doing some exercises few days ago, I saw similar questions, but clearer, like this: express $$\sqrt{28-2\sqrt{147}}$$ in the form of $$\sqrt{x}-\sqrt{y}$$. I can still handle this. But that one, I failed :(

14. satellite73

multiplying by the conjugate give $$\frac{2}{\sqrt{5+\sqrt{21}}}$$ think

15. myininaya

any examples anywhere?

16. Callisto

example?!

17. myininaya

You want to show $\sqrt{5-\sqrt{21}} \text{ equals } \frac{\sqrt{7}-\sqrt{3}}{2} ?$ was just wondering if you have an example for writing that one thing in that other form

18. myininaya

oops sqrt(2) on bottom

19. myininaya

how did you get that identity?

20. Callisto

From experience...

21. myininaya

lol great experience I want to prove that identity lol

22. Callisto

First, $$(\sqrt{a} - \sqrt{b})^2$$ = a + b -2\sqrt{ab} For a>b $\sqrt{5-\sqrt{21}} = \sqrt{5-2\sqrt{\frac{21}{4}}}$ Now, a+b = 5 => a=5-b ab = 21/4 (5-b)b = 21/4 -4b^2 + 20b - 21 =0 b=1.5 or b =3.5 (rejected) a = 5 - 1.5 = 3.5 So, it is $$\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}$$ Does that make sense?

23. myininaya

ok i see that identity :)

24. myininaya

that one is easy to prove

25. myininaya

maybe because it is an actual identity, right? :p

26. Callisto

Perfect square is perfect :)

27. Callisto

Does that make sense? Apart from the latex fail...

28. myininaya

very interesting i wouldn't have thought of that

29. Callisto

I'm going to post the link to satellite73's post then. He doesn't even come to check..

30. myininaya

Great work @Callisto :)

31. Callisto

And thank you for all your time!!!!