# How do you solve x+1/x=7?

May 25, 2018

color(red)(x=3.85 or 0.15

#### Explanation:

$x + \frac{1}{x} = 7$

or, $x + \frac{1}{x} - 7 = 0$

or, $\frac{{x}^{2} + 1 - 7 x}{x} = 0$

or, ${x}^{2} - 7 x + 1 = 0$

or, $x = \frac{- \left(- 7\right) \pm \sqrt{\left({7}^{2}\right) - 4 \cdot 1}}{2 \cdot 1} = \frac{7 \pm \sqrt{45}}{2} = \frac{7 \pm 3 \sqrt{5}}{2}$

$\therefore x = 3.5 + 1.5 \sqrt{5} \mathmr{and} 3.5 - 1.5 \sqrt{5}$

or, $x = 3.5 + 3.35 \mathmr{and} 3.5 - 3.35$

or, $x = 3.85 \mathmr{and} 0.15$

May 25, 2018

$x = \frac{7}{2} \pm \frac{3}{2} \sqrt{5}$

#### Explanation:

$\text{multiply through by } x$

${x}^{2} + 1 = 7 x$

"arrange in "color(blue)"standard form ";ax^2+bx+c=0

$\text{subtract "7x" from both sides}$

${x}^{2} - 7 x + 1 = 0$

$\text{check the value of the "color(blue)"discriminant}$

•color(white)(x)Delta=b^2-4ac

$\text{with "a=1,b=-7" and } c = 1$

${b}^{2} - 4 a c = {\left(- 7\right)}^{2} - \left(4 \times 1 \times 1\right) = 45$

$\text{thus the roots are real and irrational}$

$\text{solve using the "color(blue)"quadratic formula}$

•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)

$x = \frac{7 \pm \sqrt{45}}{2} = \frac{7 \pm 3 \sqrt{5}}{2}$

$x = \frac{7}{2} \pm \frac{3}{2} \sqrt{5} \leftarrow \textcolor{red}{\text{exact solutions}}$