Odpowiedź:
Dla x = -2 lub x = 6
Szczegółowe wyjaśnienie:
[tex]a_1 = x-2\\a_2 = x+2\\a_3 = 2x+4=2(x+2)\\x = ?[/tex]
Z własności ciągu geometrycznego:
[tex]a_2^{2} = a_1\cdot a_3\\\\(x+2)^{2} = (x-2)\cdot2(x+2)\\\\x^{2}+4x+4 = 2(x^{2}-4)\\\\x^{2}+4x+4 = 2x^{2}-8\\\\x^{2}-2x^{2}+4x+4+8 = 0\\\\-x^{2}+4x+12 = 0 \ /:(-1)\\\\x^{2}-4x-12 = 0\\\\\Delta = b^{2}-4ac = (-4)^{2}-4\cdot1\cdot(-12) = 16 + 48 = 64\\\\\sqrt{\Delta} =\sqrt{64} = 8\\\\[/tex]
[tex]x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{4-8}{2} = \frac{-4}{2} = -2\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{4+8}{2} = \frac{12}{2} = 6[/tex]