# Thread: An expression with square roots

1. ## An expression with square roots

How would you calculate

$(\sqrt {2} - \sqrt {5}) \cdot \sqrt{ 7 + 2 \sqrt {10}}$ ?

2. Hello, p.numminen!

How would you calculate: . $(\sqrt {2} - \sqrt {5}) \cdot \sqrt{ 7 + 2 \sqrt {10}}$ ?

There is a very sneaky trick we can pull . . .

First, we notice that: . $7 + 2\sqrt{10}$ .just happens to equal $\left(\sqrt{2} + \sqrt{5}\right)^2$. .
Check it out!

Then: . $\sqrt{7 + 2\sqrt{10}} \:=\:\sqrt{\left(\sqrt{2} + \sqrt{5}\right)^2} \:=\:\sqrt{2} + \sqrt{5}$

So the problem becomes: . $\left(\sqrt{2} - \sqrt{5}\right)\left(\sqrt{2} + \sqrt{5}\right) \;=\;\left(\sqrt{2}\right)^2 - \left(\sqrt{5}\right)^2 \;\;=\;2 - 5 \;\;=\;\;\boxed{-3}$

3. We can simplify $\sqrt{7+2\sqrt{10}}$ by simultaneous equations as well.

Let $\sqrt{7+2\sqrt{10}}=A+B\sqrt{10}$

Squaring, we get $7+2\sqrt{10}=A^2+10B^2+2AB\sqrt{10}$

Now by matching up coefficients, we get two equations:

$A^2+10B^2=7$ ...[1]

$2AB=2 \implies AB=1$ ...[2]

$B=\frac{1}{A}$ ...[2']

Now substituting into [1]:

$A^2+10\left(\frac{1}{A} \right)^2=7$

$A^4+10=7A^2$

$A^4-7A^2+10=0$

Now we make the substitution $u=A^2$, then it becomes:

$u^2-7u+10=0$

$(u-2)(u-5)=0$

$u=5$ or $u=2$

Then $A^2=5$ or $A^2=2$

So $A=\pm\sqrt{5}$ or $A=\pm \sqrt{2}$

So the solutions are $\{(A,B)\sqrt{5},\frac{1}{\sqrt{5}}),(-\sqrt{5},-\frac{1}{\sqrt{5}}),(\sqrt{2},\frac{1}{\sqrt{2}}), (-\sqrt{2},-\frac{1}{\sqrt{2}})\}" alt="\{(A,B)\sqrt{5},\frac{1}{\sqrt{5}}),(-\sqrt{5},-\frac{1}{\sqrt{5}}),(\sqrt{2},\frac{1}{\sqrt{2}}), (-\sqrt{2},-\frac{1}{\sqrt{2}})\}" />

Let's try the first one; then we have:

$\sqrt{7+2\sqrt{10}}=\sqrt{5}+\frac{1}{\sqrt{5}} \cdot \sqrt{10}=\sqrt{5}+\sqrt{2}$

You can check this by equating both sides as Soroban did. Note that $A=\sqrt{2}$, $B=\frac{1}{\sqrt{2}}$ also works, but the other two solutions Do Not. They are the negative solutions. That's why you have to check the final result to see if it fits!

Of course it then follows that $(\sqrt{2}-\sqrt{5})(\sqrt{2}+\sqrt{5})=2-5=-3$

4. Sometimes, one try to guess what's the expression which powered to 2 gives us the result.

Note that

$\sqrt{7+2\sqrt{10}}=\sqrt{\left(2+2\sqrt{10}+5\rig ht)}$

Now, you can see the expression $(a+b)^2$ which is hidden there.

Of course, always remember that $\sqrt{x^2}=|x|$