Compare $10.1$ and $5+\sqrt{26}$

Question

How would you compare $10.1$ and $5+\sqrt{26}$?

If we square, $102.01$ $?$ $51+10\sqrt{26}$...

score 6 · Accepted Answer · edited Apr 17 '20 at 14:25

6

$$10.1\leftrightarrow 5+\sqrt{26}$$ is $$5.1\leftrightarrow \sqrt{26}$$ and by squaring,

$$26.01\leftrightarrow 26.$$

edited Apr 17 '20 at 14:25

amWhy

answered Apr 17 '20 at 14:13

Sorry about my inadvertant edit. I thought I was editing the OP! I Rolled back to you last work, @Yves. – amWhy Apr 17 '20 at 14:26
@amWhy: thanks for the notice. Don't worry. – Apr 17 '20 at 14:55

score 1 · Answer 2 · answered Apr 17 '20 at 14:19

1

By Bernoulli $$5+\sqrt{26}=5+5\sqrt{1+\frac{1}{25}}<5+5\left(1+\frac{1}{2}\cdot\frac{1}{25}\right)=10.1.$$

answered Apr 17 '20 at 14:19

Michael Rozenberg

score 1 · Answer 3 · answered Apr 17 '20 at 15:11

1

Here is another take that does not depend on approximating $\sqrt{26}$.

$5+\sqrt{26}$ is a root of $f(x)=x^2 - 10 x - 1$. This function is increasing for $x > 5$.

Now $f(10.1)=0.01>0=f(5+\sqrt{26})$. Therefore, $10.1 > 5+\sqrt{26}$.

answered Apr 17 '20 at 15:11

lhf

3 Answers3