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प्रश्न
Solve for x:
`1/x - 1/(x-2)=3`, x ≠ 0, 2
उत्तर
We have been given,
`1/x - 1/(x-2)=3`, x ≠ 0, 2
Now we solve the above equation as follows,
`((x-2)-x)/((x-2)(x))=3`
`(-2)/(x^2-2x)=3`
-2 = 3x2 - 6x
3x2 - 6x + 2 = 0
Now we also know that for an equation ax2 + bx + c = 0, the discriminant is given by the following equation:
D = b2 - 4ac
Now, according to the equation given to us, we have, a = 3, b = -6 and c = 2.
Therefore, the discriminant is given as,
D = (-6)2 - 4(3)(2)
= 36 - 24
= 12
Now, the roots of an equation is given by the following equation,
`x=(-"b"+-sqrt"D")/(2"a")`
Therefore, the roots of the equation are given as follows,
`x=(-(-6)+-sqrt12)/(2(3))`
`=(6+-2sqrt3)/6`
`=(3+-sqrt3)/3`
Now we solve both cases for the two values of x. So, we have,
`x=(3+sqrt3)/3`
Also,
`x=(3-sqrt3)/3`
Therefore, the value of `x=(3+sqrt3)/3`, `(3-sqrt3)/3`
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