How would you show that $\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$?

Question

I've recently seen a Highschool problem and I was wondering, how would you show that

$$\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$$

Thank you for your time,

Answer 1

Both numbers are positive, so the equality is true if, and only if its squares are equal. By squaring both sides we get

$$14+4\sqrt{10}-2\sqrt{(14+4\sqrt10)(14-4\sqrt{10})}+14-4\sqrt{10} = 4^2$$

$$14-2\sqrt{(14^2-16 \cdot 10)}+14 = 16$$

$$28-12 = 16$$

$$16 = 16$$

Answer 2

$$\begin{align}\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} &= \sqrt{\left(\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}}\right)^2} \\ &= \sqrt{14 + 4\sqrt{10} + 14 - 4\sqrt{10} - 2\sqrt{(14 + 4\sqrt{10})(14-4\sqrt{10})}} \\ &= \sqrt{28 - 2\sqrt{14^2 - 4^2\cdot 10}} \\ &= \sqrt{28 - 2\cdot \sqrt{196 - 160}} \\ &= \sqrt{28 - 2\sqrt{36}} \\ &= \sqrt{28 - 12} \\ &= \sqrt{16} \\ &= 4\end{align}$$

In the above, you're applying $(a-b)^2 = a^2 + b^2 - 2ab$ and $(a+b)(a-b) = a^2 - b^2$.

Answer 3

As an alternative way note that $$\eqalign{\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} &= \sqrt{\sqrt{10}^2+2\cdot2\sqrt{10}+2^2} - \sqrt{\sqrt{10}^2-2\cdot2\sqrt{10}+2^2}},$$ then use the identity $a^2\pm2ab+b^2=(a\pm b)^2$, and simplify.

Answer 4

Since $14+4\sqrt{10}\gt14-4\sqrt{10}\gt0$, the difference of the square roots is equal to something positive, i.e.,

$$\sqrt{14+4\sqrt{10}}-\sqrt{14-4\sqrt{10}}=u\gt0$$

If we square both sides, we have

$$\begin{align} u^2&=(14+4\sqrt{10})-2\sqrt{14+4\sqrt{10}}\sqrt{14-4\sqrt{10}}+(14-4\sqrt{10})\\ &=28-2\sqrt{14^2-4^2\cdot10}\\ &=28-2\sqrt{36}\\ &=16 \end{align}$$

Since $u\gt0$ was already established, we can conclude $u=4$.

Answer 5

Denote $\sqrt{14+4\sqrt{10}}=a, \sqrt{14-4\sqrt{10}}=b$, then $$a^2+b^2=28, ab=6$$ So $(a-b)^2=16$ and $a-b=4$.

Answer 6

$$\sqrt{ 14 \pm 4 \sqrt{10}} = \sqrt{10} \pm 2 $$ since $\sqrt{10} \pm 2>0 $ and $(\sqrt{10} \pm 2)^2 = 14 \pm 4 \sqrt{10} $.

Answer 7

This was quite easy sum , here I used the general solution for (a+2b^1/2)^1/2=x^1/2 +y^1/2