Algebra Equation

Algebra Level 2

Let a = 5 + 1 5 − 1 a = \sqrt{ \dfrac{\sqrt5 + 1}{\sqrt5 - 1} } . Find the value of 5 a 2 − 5 a − 1 5a^2 - 5a - 1 .

5 3 4 0

Enigmatic Böy by Enigmatic Böy , 28, India

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2 solutions

Ikkyu San
May 12, 2016

a = 5 + 1 5 − 1 = ( 5 + 1 ) ( 5 + 1 ) ( 5 − 1 ) ( 5 + 1 ) = ( 5 + 1 ) 2 5 − 1 = 5 + 1 2 a=\sqrt{\dfrac{\sqrt5+1}{\sqrt5-1}}=\sqrt{\dfrac{(\sqrt5+1)(\sqrt5+1)}{(\sqrt5-1)(\sqrt5+1)}}=\sqrt{\dfrac{(\sqrt5+1)^2}{5-1}}=\dfrac{\sqrt5+1}2

but 5 + 1 2 \dfrac{\sqrt5+1}2 is a root of quadratic equation a 2 − a − 1 = 0 a^2-a-1=0 . Thus, 5 a 2 − 5 a − 5 = 0 ⟹ 5 a 2 − 5 a − 1 = 0 + 4 = 4 5a^2-5a-5=0\implies5a^2-5a-1=0+4=\boxed4

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Hung Woei Neoh
May 12, 2016

a = 5 + 1 5 − 1 = ( 5 + 1 ) ( 5 + 1 ) ( 5 − 1 ) ( 5 + 1 ) = ( 5 + 1 ) 2 5 − 1 = 5 + 1 2 a = \sqrt{\dfrac{\sqrt{5}+1}{\sqrt{5}-1}}\\ =\sqrt{\dfrac{(\sqrt{5} + 1)(\sqrt{5} + 1)}{(\sqrt{5}-1)(\sqrt{5}+1)}}\\ =\sqrt{\dfrac{(\sqrt{5} + 1)^2}{5-1}}\\ =\dfrac{\sqrt{5}+1}{2}

Substitute it into the expression given:

5 a 2 − 5 a − 1 = 5 a ( a − 1 ) − 1 = 5 ( 5 + 1 2 ) ( 5 + 1 2 − 1 ) − 1 = 5 ( 5 + 1 2 ) ( 5 + 1 2 − 2 2 ) − 1 = 5 2 ( 5 + 1 ) ( 5 + 1 − 2 2 ) − 1 = 5 4 ( 5 + 1 ) ( 5 − 1 ) − 1 = 5 4 ( 5 − 1 ) − 1 = 5 4 ( 4 ) − 1 = 5 − 1 = 4 5a^2 - 5a - 1\\ =5a(a-1) - 1\\ =5\left(\dfrac{\sqrt{5}+1}{2}\right)\left(\dfrac{\sqrt{5}+1}{2} - 1\right) - 1\\ =5\left(\dfrac{\sqrt{5}+1}{2}\right)\left(\dfrac{\sqrt{5}+1}{2} - \dfrac{2}{2}\right) - 1\\ =\dfrac{5}{2}\left(\sqrt{5}+1\right)\left(\dfrac{\sqrt{5}+1-2}{2}\right) - 1\\ =\dfrac{5}{4}\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right) - 1\\ =\dfrac{5}{4}\left(5-1\right) - 1\\ =\dfrac{5}{4}(4) - 1\\ =5-1\\ =\boxed{4}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

I did it in the same way...

Puneet Pinku - 5 years, 1 month ago

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