Question

sum of a number and it's reciprocal is 18 . is it possible? justify your answer​

Answer 1

Answer:

Let the number be x , so

$x + \frac{1}{x} = 18 \\ \\ {x}^{2} + 1 = 18x \\ \\ {x}^{2} - 18x + 1 = 0 \\ \\ discriminant \: d = \sqrt{ {b}^{2} - 4ac } = \sqrt{ { - 18}^{2} - 4 \times 1 \times 1} = \sqrt{320} > 0$

so it is possible