8
$\begingroup$

When I was at high school, our teacher showed us a technique to simplify square
roots like this $\sqrt{8 - 2\sqrt{7}}$ that I forgot.
It was something like 8 = 7+1; 7 = 7*1; and using them we could represent $\sqrt{8 - 2\sqrt{7}}$ in simpler form. I would be happy if you can show how it works, and how this technique is called.

  • 0
    Related: https://math.stackexchange.com/questions/757917, https://math.stackexchange.com/questions/1214527 – 2017-01-09
  • 0
    Sorry but I forgot its name. So I didn't know how to search. – 2017-01-09

7 Answers 7

6

HINT:

$$ 8=1^2+(\sqrt{7})^2 $$

and $(a-b)^2=a^2+b^2-2ab$.

3

$$\sqrt { 8-2\sqrt { 7 } } =\sqrt { { \left( \sqrt { 7 } \right) }^{ 2 }-2\sqrt { 7 } +1 } =\sqrt { { \left( \sqrt { 7 } -1 \right) }^{ 2 } } =\sqrt { 7 } -1$$

3

Maybe that the formula that you do not remember is: $$ \sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}} $$ that can easily verified ( see my answer to the similar question: Denesting a square root: $\sqrt{7 + \sqrt{14}}$) and works well when $a^2-b$ is a perfect square.

In this case we have: $$\sqrt{8 - 2\sqrt{7}}=\sqrt{8 - \sqrt{28}} \quad \Rightarrow \quad a^2-b=36$$

and, using the formula: $$ \sqrt{8 - 2\sqrt{7}}=\sqrt{7}-1 $$

2

Try making perfect square inside of that square root, as you can see that $$2\sqrt{7}=2\times1\times\sqrt{7}$$ and $$8=1+(\sqrt{7})^2$$ Then apply $$a^2+b^2-2ab=(a-b^2)$$

1

You could approach this by setting $\sqrt{8 - 2\sqrt{7}} = \sqrt a \pm \sqrt b$ for some $a,b$. Then, squaring both sides, we have:

$$8 - 2\sqrt{7} = a \pm 2\sqrt{ab} + b,$$

so that $8=a+b$ and $-2\sqrt7 = \pm 2\sqrt{ab}$.

In other words, the $\pm$ sign must be $-$, and we now have a system of two equations $\{8=a+b, 7=ab\}$ which is readily solved. Either $a=7$ and $b=1$ (which works), or $a=1$ and $b=7$ (which we reject because $1-\sqrt 7 < 0$).

0

Such square roots can be mechanically computed by a Simple Denesting Rule:

Here $\ 8-2\sqrt 7\ $ has norm $= 36.\:$ $\rm\ \color{blue}{Subtracting\ out}\,\ \sqrt{norm}\ = 6\,\ $ yields $\,\ 2-2\sqrt 7\:$

which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{4}\, =\, 2.\,\ \ \ \rm \color{brown}{Dividing\ it\ out}\,\ $ of the above yields $\ \ \ 1-\sqrt 7$

Negate it to get the positive root. Many more worked examples are in prior posts on this rule.

0

Consider this:$$\sqrt{X\pm Y}=\sqrt{\dfrac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\dfrac {X-\sqrt{X^2-Y^2}}2}\tag1$$ For $X,Y\in\mathbb{R}$ and $X>Y$. Therefore, we have $x=8,\ Y=\sqrt{28}$ so$$\sqrt{8-2\sqrt7}=\sqrt{\dfrac {8+\sqrt{64-28}}2}-\sqrt{\dfrac {8-\sqrt{64-28}}2}=\sqrt7-1\tag2$$