可以幫忙解一下數學題嗎?謝謝!!?

√( x + 1/x + 24 ) = x - √( x + 1/x + 23 )
兩邊平方, 等式仍成立:
x + 1/x + 24 = x^2 - 2x√( x + 1/x + 23 ) + x + 1/x + 23
2x√( x + 1/x + 23 ) = x^2 - 1
2√( x + 1/x + 23 ) = ( x^2 - 1 )/x = x - 1/x
兩邊平方:
4( x + 1/x + 23 ) = x^2 - 2 + 1/x^2 = ( x + 1/x )^2 - 4

令 u = x + 1/x
4( u + 23 ) = u^2 - 4
u^2 - 4u - 96 = 0
( u - 12 )( u + 8 ) = 0
u = x + 1/x = 12 , - 8 (不合, 因為x為正實數)

x + 1/x = 12
因為x為正實數, 所以 x ≠ 0 , 故等式左右皆乘以 x , 等式恆成立:
x^2 - 12x + 1 = 0
x
= [ 12 ± √( 144 - 4 ) ] / 2
= ( 12 ± 2√35 ) / 2
= 6 ± √35 ..... Ans