How to solve x + 1/x = 16

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve x + 1/x = 16 | Equatix

Question

$$x+\frac{1}{x}=16$$

Solve for x

$x=3\sqrt{7}+8\approx 15.937253933$
$x=8-3\sqrt{7}\approx 0.062746067$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.

$$xx+1=16x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}+1=16x$$

Subtract $16x$ from both sides.

$$x^{2}+1-16x=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$x^{2}-16x+1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-16$ for $b$, and $1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4}}{2}$$

Square $-16$.

$$x=\frac{-\left(-16\right)±\sqrt{256-4}}{2}$$

Add $256$ to $-4$.

$$x=\frac{-\left(-16\right)±\sqrt{252}}{2}$$

Take the square root of $252$.

$$x=\frac{-\left(-16\right)±6\sqrt{7}}{2}$$

The opposite of $-16$ is $16$.

$$x=\frac{16±6\sqrt{7}}{2}$$

Now solve the equation $x=\frac{16±6\sqrt{7}}{2}$ when $±$ is plus. Add $16$ to $6\sqrt{7}$.

$$x=\frac{6\sqrt{7}+16}{2}$$

Divide $16+6\sqrt{7}$ by $2$.

$$x=3\sqrt{7}+8$$

Now solve the equation $x=\frac{16±6\sqrt{7}}{2}$ when $±$ is minus. Subtract $6\sqrt{7}$ from $16$.

$$x=\frac{16-6\sqrt{7}}{2}$$

Divide $16-6\sqrt{7}$ by $2$.

$$x=8-3\sqrt{7}$$

The equation is now solved.

$$x=3\sqrt{7}+8$$ $$x=8-3\sqrt{7}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.

$$xx+1=16x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}+1=16x$$

Subtract $16x$ from both sides.

$$x^{2}+1-16x=0$$

Subtract $1$ from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-16x=-1$$

Divide $-16$, the coefficient of the $x$ term, by $2$ to get $-8$. Then add the square of $-8$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-16x+\left(-8\right)^{2}=-1+\left(-8\right)^{2}$$

Square $-8$.

$$x^{2}-16x+64=-1+64$$

Add $-1$ to $64$.

$$x^{2}-16x+64=63$$

Factor $x^{2}-16x+64$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-8\right)^{2}=63$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-8\right)^{2}}=\sqrt{63}$$

Simplify.

$$x-8=3\sqrt{7}$$ $$x-8=-3\sqrt{7}$$

Add $8$ to both sides of the equation.

$$x=3\sqrt{7}+8$$ $$x=8-3\sqrt{7}$$

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Example

Question

Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square