I'm trying to find a way to simplify $\sqrt{5+\sqrt{24}}$. I know that this expression is equivalent to $\sqrt{2}+\sqrt{3}$ because they are both roots of the equation: $x^4-10x^2+1$ (and the decimal equivalents are the same using a calculator). My question is , how does one get from $\sqrt{5+\sqrt{24}}$ to $\sqrt{2}+\sqrt{3}$ algebraically?
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$$(\sqrt2)^2+(\sqrt3)^2+2(\sqrt2)(\sqrt3)=?$$ – lab bhattacharjee Jul 19 '16 at 16:40
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This post may be very helpful for you: http://math.stackexchange.com/questions/1863684/fully-simplify-sqrt-14-8-sqrt-3/1863694?noredirect=1#comment3818001_1863694 – imranfat Jul 19 '16 at 16:44
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More generally, $(\sqrt{a}+\sqrt{b})^2 = a+b+2\sqrt{ab} = a + b + \sqrt{4ab}$, so you look for ways to write $24 = 4ab$ with $a+b=5$...
Robert Israel
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+1: But you could also set up a system of equations like $5=a+b$ and $ab=6$. This is equivalent to a quadratic equation, which can be solved without guessing. – MrYouMath Jul 19 '16 at 17:05
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$$\sqrt{5+\sqrt{24}}=\sqrt { 5+2\sqrt { 6 } } =\sqrt { 5+2\sqrt { 2 } \sqrt { 3 } } =\sqrt { { \left( \sqrt { 2 } +\sqrt { 3 } \right) }^{ 2 } } =\sqrt { 2 } +\sqrt { 3 } $$
Another way $$ \sqrt { 5+\sqrt { 24 } } =\sqrt { a } +\sqrt { b } \\ 5+\sqrt { 24 } =a+b+2\sqrt { ab } \\ \\ \sqrt { 24 } =2\sqrt { ab } \Rightarrow \sqrt { 6 } =\sqrt { ab } \\ \begin{cases} a+b=5 \\ \sqrt { 6 } =\sqrt { ab } \end{cases}$$
haqnatural
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If you don't want guess-and-check or hope-you-see-it approaches, suppose that $a+\sqrt{b} = \sqrt{5+\sqrt{24}}$ for some $a,b$. (Note that $a$ could be a square root.. it often works out where one of the two terms is an integer and the other is a root - which is why I wrote it in this way.) Squaring both expressions gives us
$$a^2+b + 2a\sqrt{b} = 5+\sqrt{24}.$$
Thus $a^2 + b = 5$ and $2a\sqrt{b} = \sqrt{24}$ or $4a^2b = 24.$ Thus $a^2 = 5-b$ and substituting into the other expression gives
$$4(5-b)b = 24 \Longrightarrow (5-b)b = 6 \Longrightarrow b^2-5b+6 = 0.$$
You will get two solutions to this. I will let you take it from here.
Cameron Williams
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As I explained here. there is a very simple formula for denesting such radicals, namely
Simple Denesting Rule $\rm\ \ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
$\ 5+\sqrt{24}\ $ has norm $= 1.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 1\,\ $ yields $\,\ 4+\sqrt{24}\:$
with $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{8}.\ \ \ \rm \color{brown}{Dividing\ this\ out}\ $ of the above we obtain $\,\ \sqrt 2 + \sqrt 3$
Remark $\ $ Many more worked examples are in prior posts on this denesting rule.
Bill Dubuque
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