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\displaystyle\\\left \{ {{√(x^2+2x+6)+x^2=√(2x+2) -x+3 } \atop {x=?\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }} \right.

1 Answer

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Answer:

x = 1

Explanation:

You want the solution to the equation √(x²+2x+6) +x² = √(2x+2) -x +3.

Graphing calculator

We find a graphing calculator to be a useful tool for finding the solutions to equations like this. It shows the solution to be x = 1.

Solved in the usual way, this would resolve to an 8th degree equation with one integer factor: (x -1). The remaining 7th degree polynomial factor has 3 real roots and 4 complex roots. All of these 7 irrational solutions are extraneous. (They do not satisfy the original equation.)

The solution is x = 1.

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Additional comment

Our approach to removing the radicals is to ...

  • subtract x², and square both sides
  • subtract 11 -4x -5x² +2x³ +x⁴, and square both sides
  • subtract the right side, leaving
    -47 -84x +64x² +116x³ -2x⁴ -44x⁵ -8x⁶ +4x⁷ +x⁸ = 0

The usual approach to looking for roots is to use Descartes' rule of signs and the Rational Root theorem.

You can determine there are 3 or 1 positive real roots, and 5, 3, or 1 negative real roots. The only rational real roots will be ±1 or ±47, and we know that ±47 can't work. (47^8 will not be balanced by any of the other terms in the polynomial.)

You can see already that the graphing calculator approach is much preferred.

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Solve \displaystyle\\\left \{ {{√(x^2+2x+6)+x^2=√(2x+2) -x+3 } \atop {x=?\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }} \right-example-1
User AperioOculus
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