Question

solve for x: x + 1/x = 3, x not equal to 0

Answer 1

SOLUTION

TO SOLVE

$\displaystyle \sf{x + \frac{1}{x} = 3 } \: \: \: \: \: ( \: x \ne \: 0)$

CONCEPT TO BE IMPLEMENTED

A general equation of quadratic equation is

$a {x}^{2} + bx + c = 0$

Now one of the way to solve this equation is by SRIDHAR ACHARYYA formula

For any quadratic equation

$a {x}^{2} + bx + c = 0$

The roots are given by

$\displaystyle \: x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

EVALUATION

Here the given equation is

$\displaystyle \sf{x + \frac{1}{x} = 3 }$

We simplify it as below :

$\implies \displaystyle \sf{ \frac{ {x}^{2} + 1 }{x} = 3 }$

$\implies \displaystyle \sf{ {x}^{2} + 1 = 3x }$

$\implies \displaystyle \sf{ {x}^{2} - 3x+ 1 = 0 }$

Which is of the form

$\displaystyle \sf{ a{x}^{2} + bx+ c = 0 }$

Where a = 1, b = - 3, c = 1

Hence by the SRIDHAR ACHARYYA formula we get

$\displaystyle \sf{ x = \frac{ - ( - 3) \: \pm \: \sqrt{ {( - 3)}^{2} - 4.1.1 } }{2 \times 1} \: }$

$\implies \displaystyle \sf{ x = \frac{ 3 \: \pm \: \sqrt{ 9 - 4 } }{2 } \: }$

$\implies \displaystyle \sf{ x = \frac{ 3 \: \pm \: \sqrt{ 5 } }{2 } \: }$

FINAL ANSWER

The required solution is

$\displaystyle \sf{ x = \frac{ 3 \: - \: \sqrt{ 5 } }{2 } \: \: ,\: \: \frac{ 3 \: + \: \sqrt{ 5 } }{2 }}$

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