![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
If x ＜ 0 and X & # 178; = 1 / 49, then x=
x²=1/49
It shows that x is the square root of 1 / 49,
Then x should be plus or minus 1 / 7
Because of X
36x & # 178; - 1 = 0 to solve the equation
36X2=1
The first three descending solutions of quadratic equation 36x & # 178; - 1 = 0
36x²-1=0
(6x-1)(6x+1)=0
6x-1=0 6x+1=0
∴x=±1/6
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