N is not no

Algebra Level 2

N = 5 + 2 + 5 − 2 5 + 1 − 3 − 2 2 = ? \large N = \dfrac{\sqrt{\sqrt{5} +2} + \sqrt{\sqrt{5} -2}}{\sqrt{\sqrt{5} +1}}-\sqrt{3-2\sqrt{2}} =\, ?

5 2 \frac{\sqrt{5}}{2} 1 2 5 + 1 \frac{2}{\sqrt{\sqrt{5}}+1} 2 2 − 1 2\sqrt{2} - 1

Dhruv Tyagi by Dhruv Tyagi , 19, India

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2 solutions

Dhruv Tyagi
May 6, 2016

N = N = 5 + 2 + 5 − 2 5 + 1 \frac{\sqrt{\sqrt{5} +2} + \sqrt{\sqrt{5} -2}}{\sqrt{\sqrt{5} +1}} - 3 − 2 2 \sqrt{3-2\sqrt{2}}

N + 3 − 2 2 = N+ \sqrt{3-2\sqrt{2}} = 5 + 2 + 5 − 2 5 + 1 \frac{\sqrt{\sqrt{5} +2} + \sqrt{\sqrt{5} -2}}{\sqrt{\sqrt{5} +1}}

Squaring both sides: ( N + 3 − 2 2 ) 2 = (N+ \sqrt{3-2\sqrt{2}})^{2} = ( 5 + 2 + 5 − 2 5 + 1 ) 2 (\frac{\sqrt{\sqrt{5} +2} + \sqrt{\sqrt{5} -2}}{\sqrt{\sqrt{5} +1}})^{2}

On Solving: ( N + 3 − 2 2 ) 2 = (N+ \sqrt{3-2\sqrt{2}})^{2} = 2 2

N + 3 − 2 2 = N+ \sqrt{3-2\sqrt{2}} = 2 \sqrt{2}

N − 2 = N-\sqrt{2} = − 3 − 2 2 -\sqrt{3-2\sqrt{2}}

Squaring both sides: ( N − 2 ) 2 = (N-\sqrt{2})^{2} = ( − 3 − 2 2 ) 2 (-\sqrt{3-2\sqrt{2}})^{2}

( N ) 2 − 2 ( N ) ( 2 ) + 2 = (N)^{2}-2(N)(\sqrt{2})+2 = 3 − 2 2 3-2\sqrt{2}

( N ) 2 − 2 ( N ) ( 2 ) = (N)^{2}-2(N)(\sqrt{2}) = 3 − 2 2 − 2 3-2\sqrt{2} - 2

( N ) 2 − 2 ( N ) ( 2 ) = (N)^{2}-2(N)(\sqrt{2}) = 1 − 2 2 1-2\sqrt{2}

If we substitute N by 1; we notice that N = 1 is the solution of this equation. But still:

( N ) 2 − 2 ( N ) ( 2 ) = (N)^{2}-2(N)(\sqrt{2}) = 1 − 2 2 1-2\sqrt{2}

( N ) 2 − 2 ( N ) ( 2 ) − 1 + 2 2 = (N)^{2}-2(N)(\sqrt{2})-1 + 2\sqrt{2} = 0 0

( N ) 2 − 1 − 2 ( N ) ( 2 ) + 2 2 = (N)^{2}-1-2(N)(\sqrt{2}) + 2\sqrt{2} = 0 0

( N + 1 ) ( N − 1 ) − 2 2 ( N − 1 ) = (N+1)(N-1) - 2\sqrt{2}(N-1) = 0 0

( N − 1 ) ( N + 1 − 2 2 ) = 0 (N-1)(N+1- 2\sqrt{2}) = 0

N − 1 = 0 ⇒ N-1=0 \Rightarrow N = 1 N=1 which satisfies the equation

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How do you get (N+root over(3-2 root2))^2=2

সামিন সালেক - 5 years, 1 month ago

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( 5 + 2 + 5 − 2 5 + 1 ) 2 (\frac{\sqrt{\sqrt{5} +2} + \sqrt{\sqrt{5} -2}}{\sqrt{\sqrt{5} +1}})^{2}

( 5 + 2 ) 2 + 2 ( 5 + 2 ) ( 5 − 2 ) + ( 5 − 2 ) 2 ( 5 + 1 ) 2 \frac{(\sqrt{\sqrt{5} + 2})^{2} + 2\sqrt{(\sqrt{5} + 2)(\sqrt{5} - 2)}+(\sqrt{\sqrt{5} - 2})^{2}}{(\sqrt{\sqrt{5} + 1})^{2}} On using identity ( a + b ) 2 (a+b)^{2}

5 + 2 + 2 ( 5 + 2 ) ( 5 − 2 ) + 5 − 2 5 + 1 \frac{\sqrt{5} + 2 + 2\sqrt{(\sqrt{5} + 2)(\sqrt{5} - 2)}+\sqrt{5} - 2}{\sqrt{5} + 1}

5 + 2 + 2 ( 5 ) 2 − 2 2 + 5 − 2 5 + 1 \frac{\sqrt{5} + 2 + 2\sqrt{(\sqrt{5})^{2}- 2^{2}}+\sqrt{5} - 2}{\sqrt{5} + 1} On using identity ( a + b ) ( a − b ) (a+b)(a-b)

5 + 2 + 2 5 − 4 + 5 − 2 5 + 1 \frac{\sqrt{5} + 2 + 2\sqrt{5-4}+\sqrt{5} - 2}{\sqrt{5} + 1}

5 + 2 + 2 1 + 5 − 2 5 + 1 \frac{\sqrt{5} + 2 + 2\sqrt{1}+\sqrt{5} - 2}{\sqrt{5} + 1}

5 + 2 + 2 + 5 − 2 5 + 1 \frac{\sqrt{5} + 2 + 2+\sqrt{5} - 2}{\sqrt{5} + 1}

5 + 2 + 5 5 + 1 \frac{\sqrt{5} + 2+\sqrt{5}}{\sqrt{5} + 1}

2 5 + 2 5 + 1 \frac{2\sqrt{5} + 2}{\sqrt{5} + 1}

2 ( 5 + 1 ) 5 + 1 \frac{2(\sqrt{5} + 1)}{\sqrt{5} + 1}

Now we cancel ( 5 + 1 ) (\sqrt{5} + 1) from the numerator and denominator and we get 2 \boxed{2}

Dhruv Tyagi - 5 years, 1 month ago

Your solution yields another possibility:

N = 2 2 − 1 N=2\sqrt{2} -1

You didn't show why this is not a possible answer

Hung Woei Neoh - 5 years, 1 month ago

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Try to put its value. I haven't tried though. Its one of the options too.

Dhruv Tyagi - 5 years, 1 month ago
Hung Woei Neoh
May 7, 2016

N = 5 + 2 + 5 − 2 5 + 1 − 3 − 2 2 N = \dfrac{\sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5} + 1}} - \sqrt{3 - 2\sqrt{2}}

Multiply 5 − 1 \sqrt{\sqrt{5}-1} in the fraction:

N = ( 5 − 1 ) ( 5 + 2 + 5 − 2 ) ( 5 − 1 ) ( 5 + 1 ) − 3 − 2 2 N = ( 5 − 1 ) ( 5 + 2 ) + ( 5 − 1 ) ( 5 − 2 ) ( 5 − 1 ) ( 5 + 1 ) − 3 − 2 2 N = 5 − 5 + 2 5 − 2 + 5 − 5 − 2 5 + 2 5 − 1 − 3 − 2 2 N = 1 2 ( 3 + 5 + 7 − 3 5 ) − 3 − 2 2 N = \dfrac{ \left( \sqrt{\sqrt{5}-1} \right) \left( \sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5}-2} \right)}{ \left( \sqrt{\sqrt{5}-1} \right) \left( \sqrt{\sqrt{5} + 1} \right)} - \sqrt{3 - 2\sqrt{2}}\\ N = \dfrac{ \sqrt{ \left( \sqrt{5} - 1 \right) \left( \sqrt{5} + 2 \right) } + \sqrt{ \left( \sqrt{5} - 1 \right) \left( \sqrt{5} - 2 \right) }}{\sqrt{ \left( \sqrt{5} - 1 \right) \left( \sqrt{5} + 1 \right)}} - \sqrt{3 - 2\sqrt{2}}\\ N = \dfrac{ \sqrt{ 5 - \sqrt{5} + 2\sqrt{5} - 2 } + \sqrt{ 5 - \sqrt{5} - 2\sqrt{5} + 2 }}{\sqrt{5-1}} - \sqrt{3 - 2\sqrt{2}}\\ N = \dfrac{1}{2} \left( \sqrt{3 + \sqrt{5}} + \sqrt{7 - 3\sqrt{5}} \right) - \sqrt{3 - 2\sqrt{2}}\\

Now, we want to write the following:

3 + 5 = a + b \sqrt{3+ \sqrt{5}} = \sqrt{a} + \sqrt{b} where a , b > 0 a,b > 0 . Square it:

3 + 5 = a + b + 2 a b 3+\sqrt{5} = a + b + 2\sqrt{ab}

By comparison:

3 = a + b 5 = 2 a b 5 = 4 a b b = 5 4 a 3 = a+b\\ \sqrt{5} = 2\sqrt{ab}\\ 5 = 4ab\\ b=\dfrac{5}{4a}

Substitute it in:

3 = a + 5 4 a 12 a = 4 a 2 + 5 4 a 2 − 12 a + 5 = 0 ( 2 a − 1 ) ( 2 a − 5 ) = 0 a = 1 2 , 5 2 3 = a + \dfrac{5}{4a}\\ 12a = 4a^2 + 5\\ 4a^2 - 12a + 5 = 0\\ (2a - 1)(2a - 5) = 0\\ a=\dfrac{1}{2},\; \dfrac{5}{2}

Then, b = 5 2 , 1 2 b = \dfrac{5}{2}, \; \dfrac{1}{2}

Therefore, 3 + 5 = 1 2 + 5 2 \sqrt{3+ \sqrt{5}} = \sqrt{\dfrac{1}{2}} + \sqrt{\dfrac{5}{2}}

Do the same for the following:

7 − 3 5 = c − d \sqrt{7 - 3\sqrt{5}} = \sqrt{c} - \sqrt{d} where c > d > 0 c>d>0 and

3 − 2 2 = e − f \sqrt{3 - 2\sqrt{2}} = \sqrt{e} - \sqrt{f} where e > f > 0 e>f>0

I won't show the steps for these two anymore. The method is the same, and you should get:

7 − 3 5 = 9 2 − 5 2 \sqrt{7 - 3\sqrt{5}} = \sqrt{\dfrac{9}{2}} - \sqrt{\dfrac{5}{2}}

3 − 2 2 = 2 − 1 \sqrt{3 - 2\sqrt{2}} = \sqrt{2} - \sqrt{1}

Substitute these 3 values into N N :

N = 1 2 ( 1 2 + 5 2 + 9 2 − 5 2 ) − ( 2 − 1 ) N = 1 2 ( 1 2 + 3 2 ) + 1 − 2 N = 1 2 ( 4 2 ) + 1 − 2 N = 2 + 1 − 2 N = 1 N = \dfrac{1}{2} \left( \sqrt{\dfrac{1}{2}} + \sqrt{\dfrac{5}{2}} + \sqrt{\dfrac{9}{2}} - \sqrt{\dfrac{5}{2}} \right) - \left( \sqrt{2} - \sqrt{1} \right)\\ N = \dfrac{1}{2} \left( \dfrac{1}{\sqrt{2}} + \dfrac{3}{\sqrt{2}} \right) + 1 - \sqrt{2}\\ N = \dfrac{1}{2} \left( \dfrac{4}{\sqrt{2}} \right) + 1 - \sqrt{2}\\ N = \sqrt{2} + 1 - \sqrt{2}\\ N = \boxed{1}

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