Solve x + \frac { 1 } { x } = 10 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$x + \frac { 1 } { x } = 10$$

Solve for x

$x=2\sqrt{6}+5\approx 9.898979486$
$x=5-2\sqrt{6}\approx 0.101020514$

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 $x$.

$$xx+1=10x$$

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

$$x^{2}+1=10x$$

Subtract $10x$ from both sides.

$$x^{2}+1-10x=0$$

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.

$$x^{2}-10x+1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-10$ for $b$, and $1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4}}{2}$$

Square $-10$.

$$x=\frac{-\left(-10\right)±\sqrt{100-4}}{2}$$

Add $100$ to $-4$.

$$x=\frac{-\left(-10\right)±\sqrt{96}}{2}$$

Take the square root of $96$.

$$x=\frac{-\left(-10\right)±4\sqrt{6}}{2}$$

The opposite of $-10$ is $10$.

$$x=\frac{10±4\sqrt{6}}{2}$$

Now solve the equation $x=\frac{10±4\sqrt{6}}{2}$ when $±$ is plus. Add $10$ to $4\sqrt{6}$.

$$x=\frac{4\sqrt{6}+10}{2}$$

Divide $10+4\sqrt{6}$ by $2$.

$$x=2\sqrt{6}+5$$

Now solve the equation $x=\frac{10±4\sqrt{6}}{2}$ when $±$ is minus. Subtract $4\sqrt{6}$ from $10$.

$$x=\frac{10-4\sqrt{6}}{2}$$

Divide $10-4\sqrt{6}$ by $2$.

$$x=5-2\sqrt{6}$$

The equation is now solved.

$$x=2\sqrt{6}+5$$ $$x=5-2\sqrt{6}$$

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 $x$.

$$xx+1=10x$$

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

$$x^{2}+1=10x$$

Subtract $10x$ from both sides.

$$x^{2}+1-10x=0$$

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

$$x^{2}-10x=-1$$

Divide $-10$, the coefficient of the $x$ term, by $2$ to get $-5$. Then add the square of $-5$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-1+\left(-5\right)^{2}$$

Square $-5$.

$$x^{2}-10x+25=-1+25$$

Add $-1$ to $25$.

$$x^{2}-10x+25=24$$

Factor $x^{2}-10x+25$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-5\right)^{2}=24$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{24}$$

Simplify.

$$x-5=2\sqrt{6}$$ $$x-5=-2\sqrt{6}$$

Add $5$ to both sides of the equation.

$$x=2\sqrt{6}+5$$ $$x=5-2\sqrt{6}$$