TUGAS PROBLEM SOLVING
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1) The document presents a system of 3 equations with variables X, Y, and Z: X + Y + Z = 6, X^2 + Y^2 + Z^2 = 12, and X^3 + Y^3 + Z^3 = 24. 2) It solves the system of equations by squaring the first equation, multiplying the first by the second, and multiplying the first by the third equation derived from the second to get XYZ = 8. 3) Using the formula for a cubic polynomial, it deduces the common solutions are the roots of the cubic equation x^3 - 6x^2 + 12x - 8 = 0. Therefore, the solutions are X = Y
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TUGAS PROBLEM SOLVING
- 1. PROBLEM SOLVING O L E H YUDITHSIANE MELITA TAE 131 12 061
- 2. OSN 2002 NOMOR 3 TENTUKAN SEMUA SOLUSI DARI SISTEM PERSAMAAN X + Y+ Z = 6 X族 + Y族 + Z族 = 12 X続 + Y続 + Z続 = 24
- 3. PENYELESAIAN X + Y+ Z = 6 ..........(1) X族 + Y族 + Z族 = 12....(2) X続 + Y続 + Z続 = 24....(3) KUADARATKAN PERSAMAAN (1) (X + Y+ Z)族 = 6族 X族 + Y族 + Z族 + 2(XY +XZ +YZ) = 36 XY +XZ +YZ = 12...........(4)
- 4. KALIKAN PERSAMAAN (1) DAN (2) (X + Y+ Z)(X族 + Y族 + Z族) = (6 .12) X続+Y続+Z続+XY族+XZ族+X族Y+X族Z+YZ族+Y族Z=72 XY族+XZ族+X族Y+X族Z+YZ族+Y族Z=48.........(5) KALIKAN PERSAMAAN (1) DAN (4) (X + Y+ Z)(XY +XZ +YZ) =(6.12) XY族+XZ族+X族Y+X族Z+YZ族+Y族Z+3XYZ=72 48 +3XYZ =72 XYZ = 8......(6)
- 5. RUMUS POLINOM BERDERAJAT 3 an(x-x1)(x-x2)(x-x3) =(anx - anx1) (x x2)(x - x3) =(an x2 _ anxx2 anxx1 + anx1x2) (x-x3) =an x3 - an x2x3 - an x2x2+ anxx2x3 an x2x1 + an xx1x3 + an xx1x2 - an x1x2x3 =an x3 an x2(x1+x2+x3) + an x(x1x2 + x1x3+ x2x3) - an (x1x2x3)
- 6. DARI PRSAMAAN (1) (4) DAN (6) an x3 an x2(x1+x2+x3) + an x(x1x2 + x1x3+ x2x3) - an (x1x2x3)
- 7. DARI PERSAMAAN (1)(4) DAN (6) DAPAT DISIMPULKAN BAHWA X, Y DAN Z ADALAH AKAR-AKAR PERSAMAAN x続 - 6x族 +12x - 8 = 0 sehingga (t - 2)続 = 0 Maka X=Y=Z=2 Jadi semua solusi yang memenuhi sistem Persamaan tersebut hingga X=Y=Z=2