Question

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

Answer 1

Answer:

Let the number be x , so

\begin{gathered} 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\end{gathered}

x+

x

1

=18

x

2

+1=18x

x

2

−18x+1=0

discriminantd=

b

2

−4ac

=

−18

2

−4×1×1

=

320

>0

so it is possible