A beautifully ugly equation

Algebra Level 3

If x x is a real number satisfying the following equation x 2 6 x 2 + x 2 + 2 x = ( x 2 ) 2 6 x 2 + x 2 + 1 \left| \dfrac{x-2}{6x^2+x-2}\right|+|2-x|=\dfrac{(x-2)^2}{|6x^2+x-2|}+1 If the sum of all the values of x x can be expressed in the form a b \frac{a}{b} then find the value of a + b a+b


The answer is 14.

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2 solutions

Aneesh Kundu
Jan 16, 2015

The given equation can be rewritten as x 2 6 x 2 + x 2 + x 2 = x 2 x 2 6 x 2 + x 2 + 1 \left| \dfrac{x-2}{6x^2+x-2}\right|+|x-2|=\dfrac{|x-2| \cdot |x-2|}{|6x^2+x-2|}+1 Let

x 2 = a |x-2|=a and

x 2 6 x 2 + x 2 = b \left| \dfrac{x-2}{6x^2+x-2}\right| =b

Now the expression can be rewritten in terms of a a and b b as b + a = a b + 1 \Rightarrow b+a=ab+1 ( a 1 ) ( b 1 ) = 0 \Rightarrow (a-1)(b-1)=0 a = 1 , b = 1 \Rightarrow a=1, b=1

  • Case 1

x 2 = 1 x = 1 , 3 |x-2|=1\Rightarrow x=1,3

  • Case 2

x 2 6 x 2 + x 2 = 1 \left| \dfrac{x-2}{6x^2+x-2}\right|=1

x 2 6 x 2 + x 2 = ± 1 \Rightarrow\dfrac{x-2}{6x^2+x-2}=\pm 1 x = 1 , 0 , 2 3 \Rightarrow x=-1,0, \frac{2}{3}

x = 1 , 0 , 2 3 , 1 , 3 \Rightarrow \boxed{x=-1,0, \frac{2}{3},1,3}

what a wonderful solution. Keep it up.

Sandeep Bhardwaj - 6 years, 2 months ago

Shit. I didn't only have -1. Nice solution

Aayush Patni - 6 years, 2 months ago
Chew-Seong Cheong
Jan 25, 2015

The solution can be obtained by plotting the curve as follows (if it is not considered cheating):

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