1. ## Radicals calcules

You have : ,
Calculate :

Hello dhiab
Originally Posted by dhiab
You have : ,
Calculate :
B is not a real number. Are you sure you have the question correct?

3. Originally Posted by Grandad
Hello dhiabB is not a real number. Are you sure you have the question correct?

Hello : THANK YOU

4. Hello dhiab

I'm not sure what form you want the answer in - I assume you want it as a single radical. How does this look?

Note first that $\displaystyle AB = \sqrt{(\sqrt5+2)(\sqrt5-2)}=\sqrt{5-4}=1$

Let $\displaystyle x = \sqrt{A} + \sqrt{B}$

Then $\displaystyle x^2 = A + 2\sqrt{AB} + B = A +2+B$

$\displaystyle \Rightarrow (x^2)^2 =(A+B+2)^2= A^2 +2AB + B^2 + 4(A+B) + 4$

Noting that $\displaystyle A+B = x^2 - 2$, this simplifies to

$\displaystyle (x^2)^2 = 4x^2 +2\sqrt5 - 2$

$\displaystyle \Rightarrow (x^2)^2 - 4x^2 -(2\sqrt5-2)=0$

$\displaystyle \Rightarrow x^2 = \frac{4\pm\sqrt{16 + 4(2\sqrt5-2)}}{2}$, using the quadratic formula

This then simplifies to

$\displaystyle x=\sqrt{2+\sqrt{2+2\sqrt{5}}}$

5. Hello, everyone!

Grandad is right . . . I misread the problem . . . *blush*

As Emily Litella would say: "Never mind."

6. Hello Soroban
Originally Posted by Soroban
...
Let $\displaystyle x \:=\:A + B$...
But we are asked to find $\displaystyle \sqrt{A} + \sqrt{B}$

7. Hello,
Originally Posted by Grandad

Note first that $\displaystyle AB = \sqrt{(\sqrt5+2)(\sqrt5-2)}=\sqrt{5-4}=1$

Let $\displaystyle x = \sqrt{A} + \sqrt{B}$

Then $\displaystyle x^2 = A + 2\sqrt{AB} + B = A +2+B$
... hence
$\displaystyle x=+\sqrt{A+2+B}=\sqrt{\sqrt{5}+2+\sqrt{5}-2}=\sqrt{2+2\sqrt{5}},$
there is no need for the quadratic formula.

Originally Posted by Grandad

$\displaystyle \Rightarrow (x^2)^2 =(A+B+2)^2= A^2 +2AB + B^2 + 4(A+B) + 4$

Noting that $\displaystyle A+B = x^2 - 2$, this simplifies to

$\displaystyle (x^2)^2 = 4x^2 +2\sqrt5 - 2$
I think the last equation is not correct. We have
\displaystyle \begin{aligned}(x^2)^2&=(A+B+2)^2\\ &=(A+B)^2 + 4(A+B) + 4\\ &=(2\sqrt{5})^2+4(x^2-2)+4\\ &=4x^2+20-8+4\\ (x^2)^2&=4x^2+16\end{aligned}

and the positive root of $\displaystyle X^2-4X^2-16$ is $\displaystyle \frac{4+\sqrt{16+4\times 16}}{2}=2+2\sqrt{5}$ so we get $\displaystyle x=+\sqrt{2+2\sqrt{5}}$ which matches the solution given above.

8. ## Not enough square root signs

Hello everyone

Sorry flyingsquirrel, but you haven't read the question correctly - you haven't got enough square root signs
Originally Posted by flyingsquirrel
Hello,

... hence
$\displaystyle x=+\sqrt{A+2+B}=\sqrt{\sqrt{5}+2+\sqrt{5}-2}=\sqrt{2+2\sqrt{5}},$
there is no need for the quadratic formula.

...
$\displaystyle A = \sqrt{\sqrt5+2}, B = \sqrt{\sqrt5-2}$

$\displaystyle \Rightarrow A+ B+2 = \sqrt{\sqrt5+2}+ \sqrt{\sqrt5-2}+2$

$\displaystyle \ne 2\sqrt5 +2$

It is $\displaystyle A^2 + B^2$ that equals $\displaystyle 2\sqrt5$, not $\displaystyle A + B$.

So my original answer stands.