Group Them Carefully 3

Algebra Level 1

If x x is a positive real number satisfying the equation x − 1 x = 3 x - \dfrac1x = 3 , what is the value of x + 1 x x+ \dfrac1x ?

7 \sqrt{7} 13 \sqrt{13} 19 \sqrt{19} 25 \sqrt{25}

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Chung Kevin by Chung Kevin , 45, Australia

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2 solutions

( x − 1 x ) 2 = 3 2 x 2 + 1 x 2 − 2 = 9 x 2 + 1 x 2 + 2 = 13 ( x + 1 x ) 2 = 13 x + 1 x = 13 ( x > 0 ) \begin{aligned} \left( x - \dfrac1x \right)^2 &=& 3^2 \\ x^2 + \dfrac1{x^2} - 2 &=& 9\\ x^2 + \dfrac1{x^2} +2 &=& 13 \\\left( x + \dfrac1x \right)^2 &=& 13 \\ x + \dfrac{1}{x} &=& \sqrt{13} &(x>0) \end{aligned}

16 Helpful 0 Interesting 1 Brilliant 0 Confused

Yes! This is correct too! Note that we don't have to use the quadratic formula. ;)

Chung Kevin - 5 years, 3 months ago
Akhil Bansal
Mar 16, 2016

( x + 1 x ) 2 = ( x − 1 x ) 2 + 4 \large \left(x + \dfrac{1}{x}\right)^2 = \left(x - \dfrac{1}{x}\right)^2 + 4 ( x + 1 x ) 2 = 3 2 + 4 \large \left(x + \dfrac{1}{x}\right)^2 = 3^2 + 4 x + 1 x = 13 \large x + \dfrac{1}{x} = \boxed{\sqrt{13}}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Thank you for your solution. I came up with this problem using your very first line.

Chung Kevin - 5 years, 3 months ago

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This method is helpful when we have the value of x + y x+y and x y xy and we have to find value of x , y x,y .

Akhil Bansal - 5 years, 3 months ago

i dont see how u get line 1?

Gabriel Lopez - 5 years, 2 months ago

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( x + 1 x ) 2 = x 2 + 1 x 2 + 2 x ⋅ 1 x \left(x + \dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2x\cdot \dfrac{1}{x}
( x − 1 x ) 2 = x 2 + 1 x 2 − 2 x ⋅ 1 x \left(x - \dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} - 2x\cdot\dfrac{1}{x}

Adding above 2 equations, you'll get the desired result.

Akhil Bansal - 5 years, 2 months ago

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