Advanced Math Test #1

Algebra Level 2

Calculate ( 12 x 2 + 4 x − 57 ) 2019 (12x^2+4x-57)^{2019} for x = 10 + 6 3 3 ( 3 − 1 ) 6 + 2 5 − 5 x=\dfrac{\sqrt[3]{10+6\sqrt{3}}(\sqrt{3}-1)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}} .


The answer is -1.

Tin Le by Tin Le , 15, Vietnam

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2 solutions

Chew-Seong Cheong
Oct 14, 2019

Given that x = 10 + 6 3 3 ( 3 − 1 ) 6 + 2 5 − 5 = ( 3 + 1 ) 3 3 ( 3 − 1 ) ( 1 + 5 ) 2 − 5 = ( 3 + 1 ) ( 3 − 1 ) 1 + 5 − 5 = 2 1 = 2 x = \dfrac {\sqrt[3]{10+6\sqrt 3}(\sqrt 3-1)}{\sqrt{6+2\sqrt 5}-\sqrt 5} = \dfrac {\sqrt[3]{(\sqrt 3+1)^3}(\sqrt 3-1)}{\sqrt{(1+\sqrt 5)^2}-\sqrt 5} = \dfrac {(\sqrt 3+1)(\sqrt 3-1)}{1+\sqrt 5 - \sqrt 5} = \dfrac 21 = 2 .

Therefore ( 12 x 2 + 4 x − 57 ) 2019 = ( 48 + 8 − 57 ) 2019 = ( − 1 ) 2019 = − 1 (12x^2+4x-57)^{2019} = (48+8-57)^{2019} = (-1)^{2019} = \boxed{-1} .

1 Helpful 1 Interesting 1 Brilliant 0 Confused

After simplification we get x = 2 x=2 and the given quantity = − 1 =-1

0 Helpful 0 Interesting 0 Brilliant 0 Confused

How would you simplify this

Δrchish Ray - 1 year, 8 months ago

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( 3 + 1 ) 3 = 10 + 6 3 , ( 3 + 1 ) ( 3 − 1 ) = 2 (\sqrt 3+1)^3=10+6\sqrt 3, (\sqrt 3+1)(\sqrt 3-1)=2 . 6 + 2 5 = 5 + 1 \sqrt {6+2\sqrt 5}=\sqrt 5+1 . So the numerator is 2 2 and the denominator is 1 1 , and x = 2 1 = 2 x=\dfrac{2}{1}=2

A Former Brilliant Member - 1 year, 8 months ago

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