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quadratic equation

• a^3+8b^3+a^2-4b^2
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發問:

√x-12/√x＝1 3(5^2x)+26(5^x)-9=0 更新: 1/(x-2)^2+1/x+2-6=0 更新 2: 點解1/y^2 + 1/y - 6=0會變成 1 + y - 6y^2 = 0

最佳解答:

1) Sub y = √x, then we have: y - 12/y = 1 y2 - 12 = y y2 - y - 12 = 0 (y - 4)(y + 3) = 0 y = 4 or -3 (rejected since √x must be positive) √x = 4 x = 16 2) Let y = 5x, then 3y2 + 26y - 9 = 0 (3y - 1)(y + 9) = 0 y = 1/3 or -3 (rejected since 5x must be positive) 5x = 1/3 x log 5 = - log 3 x = - log 3/log 5 2010-08-19 23:25:45 補充： 3) Let y = x - 2, then 1/y^2 + 1/y - 6=0 1 + y - 6y^2 = 0 (1 + 3y)(1 - 2y) = 0 y = -1/3 or 1/2 x - 2 = -1/3 or 1/2 x = 5/3 or 5/2 2010-08-20 08:57:10 補充： Q: 點解1/y^2 + 1/y - 6=0會變成 1 + y - 6y^2 = 0 A: 將兩邊乘上 y^2

其他解答:

√x-12/√x＝1 X-12=√X X-√X-12=0 [(√X)-4][(√X)+3]=0 √X=4 OR √X=-3(REJ) x=16 3(5^2x)+26(5^x)-9=0 [3(5^x)-1][(5^x)+9]=0 5^x = 1/3 or 5^x = -9 (rej) x log5 = log (1/3) x= -0.683(corr to 3 sig fig) 2010-08-19 23:48:08 補充： 1/(x-2)^2+1/x+2 -6=0 請問係唔係應該係 1/(x-2)^2+1/x-2-6=0 ? 2010-08-19 23:50:41 補充： if yes, 1/(x-2)^2+1/(x-2) -6=0 [(1/x-2)+3][(1/x-2)-2]=0 1/x-2 = -3 or 1/x-2 = 2 x=5/3 or x = 2.5