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(X+1) (X+2) (X+3) (X+4) = 120 |
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Answer» Rewrite the equation as:(x + 1)(x + 4)(x + 3)(x + 2) = 120Multiply the first 2 and last 2 expressions:\xa0..... (1)Let\xa0(1) becomes,(y + 4)(y + 6) = 120y2\xa0+ 10x + 24 = 120y2\xa0+ 10x - 96 = 0y2\xa0+ 16x - 6y - 96 = 0y(y + 16) - 6(y + 16) = 0(y + 16)(y - 6) = 0y = -16 or 6y2+ 5x = 16 or 6When x2\xa0+ 5x = 6x2\xa0+ 5x - 6 = 0x2\xa0+ 6x - x - 6 = 0x(x + 6) - 1(x + 6) = 0(x + 6)(x - 1) = 0x = -6, 1When x^2 + 5x = -16x^2 + 5x + 16 = 0Using quadratic These are all the possible solutions. But the only real solutions are -6 and 1. Rewrite the equation as:(x + 1)(x + 4)(x + 3)(x + 2) = 120Multiply the first 2 and last 2 expressions:{tex}(x^2 + 5x + 4)(x^2 + 5x + 6) = 120{/tex} ..... (1)Let {tex}x^2 + 5x = y{/tex}(1) becomes,(y + 4)(y + 6) = 120y2 + 10x + 24 = 120y2 + 10x - 96 = 0y2 + 16x - 6y - 96 = 0y(y + 16) - 6(y + 16) = 0(y + 16)(y - 6) = 0y = -16 or 6y2+ 5x = 16 or 6When x2 + 5x = 6x2 + 5x - 6 = 0x2 + 6x - x - 6 = 0x(x + 6) - 1(x + 6) = 0(x + 6)(x - 1) = 0x = -6, 1When x^2 + 5x = -16x^2 + 5x + 16 = 0Using quadratic formula:{tex}x = [-5 +/- sqrt (25 - 64)]/2{/tex}The solutions are, -6, 1, [-5 - sqrt (25 - 64)]/2 and [-5 + sqrt (25 - 64)]/2These are all the possible solutions. But the only real solutions are -6 and 1. |
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