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    Math Problems
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    • mathnerd_101M
      mathnerd_101 M0 M3
      last edited by

      Here's a little fun substitution problem!

      Find all real numbers x such that $$(x^2 - x - 2)^4 + (2x+1)^4 = (x^2+x-1)^4.$$

      If you see the text messages between two programmers, here are some common acronyms to know:
      imo: iteration may overload
      brb: bad recursion brb
      rofl: right-oriented frame layout

      1 Reply Last reply Reply Quote 0
      • E
        excitedarmadillo 0 M1★ M2 M3 M5★
        last edited by

        @mathnerd_101

        The answers are -1, -0.5, and 2.

        mathnerd_101M 1 Reply Last reply Reply Quote 0
        • mathnerd_101M
          mathnerd_101 M0 M3 @excitedarmadillo 0
          last edited by

          @excitedarmadillo-0 Correct! Now, what was your solution to this?

          If you see the text messages between two programmers, here are some common acronyms to know:
          imo: iteration may overload
          brb: bad recursion brb
          rofl: right-oriented frame layout

          1 Reply Last reply Reply Quote 0
          • E
            excitedarmadillo 0 M1★ M2 M3 M5★
            last edited by

            Painfully expand and simplify.

            mathnerd_101M 1 Reply Last reply Reply Quote 0
            • mathnerd_101M
              mathnerd_101 M0 M3 @excitedarmadillo 0
              last edited by

              @excitedarmadillo-0 uhh :concern: are you ok?

              ANYWAYS, here's the non-i-want-to-die-and-i'm-a-masochist solution!
              Let us assume that

              $$a = x^2 - x - 2, b = 2x+1. $$

              Thus, the equation can be written as $$a^4 + b^4 = (a+b)^4 = a^4 + 4a^3b+6a^2b^2+4ab^3+b^4.$$

              Simplifying, we get $$2ab(2a^2 + 3ab+2b^2) = 0.$$

              Assuming that a=0, we get that $$x^2-x-2 = 0,$$ thus x = -1 or x = 2, which are the solutions of the equation.

              Assume now that b=0, we get that $$x = -\frac{1}{2}.$$

              Finally, assume that ab does not equal 0, then we get $$2a^2+3ab+2b^2 = 0.$$ As this is a quadratic equation in a, the discriminant is $$-7*b^2<0.$$

              Thus, this equation has no solution, meaning that we have 3 solutions: $$-1, -\frac{1}{2}, 2 \blacksquare$$

              If you see the text messages between two programmers, here are some common acronyms to know:
              imo: iteration may overload
              brb: bad recursion brb
              rofl: right-oriented frame layout

              1 Reply Last reply Reply Quote 0
              • S
                studiousweasel 0 M2
                last edited by

                Yes very fun and totally not pain and suffering. 🙂

                1 Reply Last reply Reply Quote 0

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