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प्रश्न
Solve for x
`x+1/x=3`, x ≠ 0
उत्तर
We have been given,
`x+1/x=3`, x ≠ 0
Now, we solve the equation as follows:
`(x^2+1)/x=3`
x2 + 1 = 3x
x2 - 3x + 1 = 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 = 1, b = -3 and c = 1.
Therefore, the discriminant is given as,
D = (-3)2 - 4(1)(1)
= 9 - 4
= 5
Now, the roots of an equation is given by the following equation,
`x=(-b+-sqrtD)/(2a)`
Therefore, the roots of the equation are given as follows,
`x=(-(-3)+-sqrt5)/(2(1))`
`=(3+-sqrt5)/2`
Now we solve both cases for the two values of x. So, we have,
`x=(3+sqrt5)/2`
Also,
`x=(3-sqrt5)/2`
Therefore, the value of `x=(3+sqrt5)/2`, `(3-sqrt5)/2`
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