根号(7x^2+9x+13)+根号(7x^2-5x+13)=7x求x的值...(题中均为二次根式).

Question

根号(7x^2+9x+13)+根号(7x^2-5x+13)=7x 求x的值...(题中均为二次根式).

Answer 1

sqrt(7x^2 + 9x + 13) = 7x - sqrt(7x^2 - 5x + 13)
两边平方
7x^2 + 9x + 13 = 49x^2 + 7x^2 - 5x + 13 - 14x·sqrt(7x^2 - 5x + 13)
49x^2 - 14x = 14x·sqrt(7x^2 - 5x + 13)
显然 x = 0 不是方程的解,两边除以 7x 再平方
(7x - 2)^2 = 4(7x^2 - 5x + 13)
49x^2 - 28x + 4 = 28x^2 - 20x + 52
21x^2 - 8x - 48 = 0
(7x - 12)(3x + 4) = 0
7x 等于两个平方根之和,所以 x 不小于 0,(3x + 4) > 0
所以,x = 12/7