# Estimate a value with square roots squared without calculating

If $$n$$ is a natural number and $$n<(6+\sqrt{29})^2, find the value of $$n$$.

This problem is, of course, very easy to solve with a calculator. But I am looking for a solution that does NOT need to find the value of $$(6+\sqrt{29})^2$$. Is there a way to do that?

• This is an irrational number. How many decimals ? – Yves Daoust Jan 27 at 12:19

Without a calculator,

$$(6+\sqrt{29})^2=36+12\sqrt{29}+29=65+12\sqrt{29}=125+12(\sqrt{29}-5)=125+{48\over\sqrt{29}+5}$$

Now $$5\lt\sqrt{29}\lt6$$ implies

$$4\lt{48\over11}\lt{48\over\sqrt{29}+5}\lt{48\over10}\lt5$$

Thus

$$129\lt(6+\sqrt{29})^2\lt130$$

• Nice proof (+1) – Angelo Jan 27 at 12:25
• Why make the denominator $\sqrt{29}+5$? Is +4 or +3 fine? – Cyh1368 Jan 27 at 13:23
• @Cyh1368, $\sqrt{29}+5$ is the "conjugate" to $\sqrt{29}-5$, i.e., $(\sqrt{25}+5)(\sqrt{29}-5)=29-25=4$. – Barry Cipra Jan 27 at 14:41
• @BarryCipra But what if we used $\sqrt{29}-4$ in the first place? Then can't we just ues $\sqrt{29}+4$ as the conjugate? – Cyh1368 Jan 28 at 2:22
• @Cyh1368, ah, I see what you're getting at: How did I know to turn $65+12\sqrt{29}$ into $125+12(\sqrt{29}-5)$ rather than $113+12(\sqrt{29}-4)$? The idea is to use a $12(\sqrt{29}-a)$ that makes the numerator, $12(29-a^2)$, as small as possible. – Barry Cipra Jan 28 at 11:27

Expand the expression to get $$36+12\sqrt{29}+29=65+\sqrt{4176}$$. Now you only need to find the integer $$m$$ such that $$m<\sqrt{4176}, i.e. $$m^2<4176<(m+1)^2$$.

• It's a simple solution, but still need some time to figure out $\sqrt{4176}$ – Cyh1368 Jan 27 at 13:01
• @Cyh1368: Yes, you are right -- I prefer the other solutions to mine. (But as every programmer knows, $64^2=4096$.) – TonyK Jan 27 at 13:18
• (+1), and $65^2 = (6\times 7)\times 100+25=4225$ is a well-known trick too. – Neat Math Jan 27 at 14:12

$$5<\sqrt{29}<6$$, so $$0<6-\sqrt{29}<1$$, so $$0<(6-\sqrt{29})^2<1$$.

$$(6+\sqrt{29})^2<(6+\sqrt{29})^2+(6-\sqrt{29})^2<(6+\sqrt{29})^2+1$$.

But $$(6+\sqrt{29})^2+(6-\sqrt{29})^2=36+12\sqrt{29}+29+36-12\sqrt{29}+29=130$$.

So $$(6+\sqrt{29})^2<130<(6+\sqrt{29})^2+1$$.

So $$129<(6+\sqrt{29})^2<130$$.

Use the difference of two squares to get:

$$6+\sqrt{29} = \frac{(6+\sqrt{29})(6-\sqrt{29})}{6-\sqrt{29}} = \frac{7}{6 - \sqrt{29}}$$

and so:

$$n<\frac{49}{(6-\sqrt{29})^2}

Adding $$49$$ times this inequality with the original gives:

$$n + \frac{49}{n+1} < (6 + \sqrt{29})^2 + (6 - \sqrt{29})^2 < n+1 + \frac{49}{n}$$ $$\Rightarrow n + \frac{49}{n+1} < 130 < n+1 + \frac{49}{n}$$

From the right-hand side, the minimum value of $$n$$ which satisfies the inequality is $$129$$. But from the left-hand side, $$n$$ cannot be $$130$$ or greater. Therefore $$n = 129$$ which is the unique solution.

• Note that solving this inequality by expanding and using the quadratic formula would be horrendous. – Toby Mak Jan 27 at 12:43