Question

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Решите уравнение (х²-3)²-4(х²-3)+3=0

Answer 1

${( {x}^{2} - 3)}^{2} - 4( {x}^{2} - 3) + 3 = 0 \\ сделаем \: замену \: переменной \: \: \\ {x}^{2} - 3 = y \\ {y}^{2} - 4y + 3 = 0 \\ d = 16 - 12 = 4 \\ y1 = \frac{4 + \sqrt{4} }{2} = \frac{4 + 2}{2} = \frac{6}{2} = 3 \\ y2 = \frac{4 - 2}{2} = \frac{2}{2} = 1 \\ произведем \: обратную \: замену \\ {x}^{2} - 3 = y1 \\ {x}^{2} - 3 = 3 \\ {x}^{2} = 6 \\ x = ± \sqrt{6} \\ - - - - - \\ {x}^{2} - 3 = y2 \\ {x}^{2} - 3 =1 \\ {x}^{2} = 4 \\ x = \sqrt{4} \\ x = ±2 \\ otvet \\ x = (±2 ; \: \: ± \sqrt{6} )$