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Algebra Level 3

If 4 x 2 + 1 4 x 2 = 7 4x^2+\dfrac{1}{4x^2} = 7 , what is the largest possible value of 2 x + 1 2 x 2x+\dfrac{1}{2x} ?


The answer is 3.

Akeel Howell by Akeel Howell , 22, USA

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1 solution

Akeel Howell
Mar 25, 2017

( 2 x + 1 2 x ) 2 = 4 x 2 + 1 4 x 2 + 2 = 7 + 2 = 9 ∴ 4 x 2 + 1 4 x 2 + 2 = 9 = 3 \left(2x+\dfrac{1}{2x}\right)^2 = 4x^2+\dfrac{1}{4x^2}+2 = 7+2 = 9 \\ \therefore \sqrt{4x^2+\dfrac{1}{4x^2}+2} = \sqrt{9} = \space 3 .

Clearly, 3 3 is the largest possible value of 2 x + 1 2 x 2x+\dfrac{1}{2x} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

The notation that 9 \sqrt{9} is ± 3 \pm 3 is wrong. The equations x = a x=\sqrt{a} and x 2 = a x^2=a are completely different from each other as one of them is linear and the other one quadratic. The appropriate notation would be

f ( x ) 2 = 9 ⟹ f ( x ) = ± 3 f(x)^2 = 9 \implies f(x) = \pm 3

(here f ( x ) f(x) is 2 x + 1 2 x 2x + \dfrac{1}{2x} ).

Tapas Mazumdar - 4 years, 1 month ago

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