# Reciprocal value

How do I calculate a number x that is 9 greater than its reciprocal (1/x)?

Result

x1 =  9.10977
x2 =  -0.10977

#### Solution:

$x=9 + 1/x \ \\ x^2=9x + 1 \ \\ \ \\ x^2=9x + 1 \ \\ \ \\ x^2 -9x -1=0 \ \\ \ \\ a=1; b=-9; c=-1 \ \\ D=b^2 - 4ac=9^2 - 4\cdot 1 \cdot (-1)=85 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 9 \pm \sqrt{ 85 } }{ 2 } \ \\ x_{1,2}=4.5 \pm 4.60977222865 \ \\ x_{1}=9.10977222865=9.10977 \ \\ x_{2}=-0.109772228646 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -9.10977222865) (x +0.109772228646)=0$

Checkout calculation with our calculator of quadratic equations.

$x_{2}=(-0.1098) \doteq -0.1098=-0.10977$

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