Find the roots of 1x−1x−2=3
We have,
1x−1x−2=3
x−2−xx(x−2)=3
−2(x2−2x)=3
−2=3x2−6x
3x2−6x+2=0 ……. (1)
We know that
x=−b±√b2−4ac2a
Therefore,
x=6±√62−4×3×22×3
x=6±√36−246
x=6±√126
x=6±2√36
x=1±√33
x=1±1√3
Hence, the roots of this equation are 1+1√3,1−1√3.