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Algebra Level 4

{ x 4 + 2 x 3 − y = − 1 4 + 3 y 4 + 2 y 3 − x = − 1 4 − 3 \begin{cases}x^4+2x^3-y=-\frac{1}{4}+\sqrt{3}\\ y^4+2y^3-x=-\frac{1}{4}-\sqrt{3} \end{cases}

All the ordered pairs of real numbers that satisfy the system of equations above are ( x 1 , y 1 ) , ( x 2 , y 2 ) , . . . ( x n , y n ) (x_1,y_1),(x_2,y_2),...(x_n,y_n) . Find the value of x 1 + x 2 + . . . + x n + y 1 + . . . y n x_1+x_2+...+x_n+y_1+...y_n correct upto two decimal places.

  • If you think there are infinite solutions then answer 777 and if you think no real solutions answer 666.
  • Ordered pair means (11,12),(12,11) are considered different.


The answer is -1.

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Ravi Dwivedi by Ravi Dwivedi , 24, India

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1 solution

Ravi Dwivedi
Jul 13, 2015

Adding the equations we get ( x 4 + 2 x 3 − x ) + ( y 4 + 2 y 3 − y ) = − 1 2 (x^4+2x^3-x)+(y^4+2y^3-y)=-\frac{1}{2} ( x 2 + x − 1 2 ) 2 + ( y 2 + y − 1 2 ) 2 = 0 (x^2+x-\frac{1}{2})^2+(y^2+y-\frac{1}{2})^2=0 Equality holds only if ( x 2 + x − 1 2 ) 2 = ( y 2 + y − 1 2 ) 2 = 0 (x^2+x-\frac{1}{2})^2=(y^2+y-\frac{1}{2})^2=0 On solving the equations we get that

{ x , y } ⊂ { − 1 2 − 3 2 , − 1 2 + 3 2 } \{x,y\} \subset \{-\frac{1}{2}-\frac{\sqrt{3}}{2},-\frac{1}{2}+\frac{\sqrt{3}}{2}\}

Notice that x ≠ y x \neq y because if x = y x=y then from original equations − 1 4 + 3 = − 1 4 − 3 -\frac{1}{4}+\sqrt{3}=-\frac{1}{4}-\sqrt{3} which is not true.

So two ordered pairs are there ( x , y ) = ( − 1 2 − 3 2 , − 1 2 + 3 2 ) (x,y)=(-\frac{1}{2}-\frac{\sqrt{3}}{2},-\frac{1}{2}+\frac{\sqrt{3}}{2}) and ( − 1 2 + 3 2 , − 1 2 − 3 2 ) (-\frac{1}{2}+\frac{\sqrt{3}}{2},-\frac{1}{2}-\frac{\sqrt{3}}{2})

Answer is − 1 2 − 1 2 − 1 2 − 1 2 = − 2 -\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-2

Edit: Note that the first ordered pair doesn't satisfy the original system of equations. Hence the answer is only − 1 - 1 .
When manipulating a system of non-linear equations, we might introduce solutions even with simple operations like addition and subtraction.

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

After solving a combined system of equations, we have to verify that these are indeed solutions to the original system. Sometimes we introduce extraneous solutions.

Your solution is -2, but the answer showed is "The correct answer is -1.". I guess the "correct answer" in the system is wrong, isn't it?

Ricardo Takayama - 5 years, 8 months ago

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I put the answer as -2 but I don't know why Brilliant changed it to "-1". Also the feedback by challenge master is already given.Do u agree with my solution?

Ravi Dwivedi - 5 years, 8 months ago

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Yes, I do! =)

Ricardo Takayama - 5 years, 8 months ago

The answer is − 1 -1 , because only x = − 1 2 + 3 2 x=-\dfrac 12+\dfrac{\sqrt{3}}{2} and y = − 1 2 − 3 2 y=-\dfrac12-\dfrac{\sqrt{3}}{2} is a solution to any of the original equations.

Laurent Shorts - 5 years, 2 months ago

Note that there are 4 choices of ++, +-, -+, --. No equal roots can work, since that violates that the above form is in +-, and not ++ or -- form due to equation symmetry. Only one of the ordered pairs can work, as if both works then the +- and -+ forms could work, which they do not. Then notice that in that pair, the sqrt(3)/2 pairs cancel out, leaving (-1/2)+(-1/2) = -1

Rachel Laubacher - 4 years ago

this is weird but what was your approach? I mean, is this a standard approach for manipulating biquadratic equations by converting into the square of a quadratic trinomial or was there something special here about the coefficients and the constant that I was unable to see while solving the problem?

Vandit Kumar - 3 years, 3 months ago

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