How to solve x+(1)/(x)=5

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)=5 | Equatix

Question

$$x+ \frac{ 1 }{ x } =5$$

Solve for x

$x = \frac{\sqrt{21} + 5}{2} \approx 4.791287847$
$x=\frac{5-\sqrt{21}}{2}\approx 0.208712153$

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 $x$.

$$xx+1=5x$$

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

$$x^{2}+1=5x$$

Subtract $5x$ from both sides.

$$x^{2}+1-5x=0$$

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.

$$x^{2}-5x+1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-5$ for $b$, and $1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4}}{2}$$

Square $-5$.

$$x=\frac{-\left(-5\right)±\sqrt{25-4}}{2}$$

Add $25$ to $-4$.

$$x=\frac{-\left(-5\right)±\sqrt{21}}{2}$$

The opposite of $-5$ is $5$.

$$x=\frac{5±\sqrt{21}}{2}$$

Now solve the equation $x=\frac{5±\sqrt{21}}{2}$ when $±$ is plus. Add $5$ to $\sqrt{21}$.

$$x=\frac{\sqrt{21}+5}{2}$$

Now solve the equation $x=\frac{5±\sqrt{21}}{2}$ when $±$ is minus. Subtract $\sqrt{21}$ from $5$.

$$x=\frac{5-\sqrt{21}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{21}+5}{2}$$ $$x=\frac{5-\sqrt{21}}{2}$$

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 $x$.

$$xx+1=5x$$

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

$$x^{2}+1=5x$$

Subtract $5x$ from both sides.

$$x^{2}+1-5x=0$$

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

$$x^{2}-5x=-1$$

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

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

Square $-\frac{5}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-1+\frac{25}{4}$$

Add $-1$ to $\frac{25}{4}$.

$$x^{2}-5x+\frac{25}{4}=\frac{21}{4}$$

Factor $x^{2}-5x+\frac{25}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{21}}{2}$$ $$x-\frac{5}{2}=-\frac{\sqrt{21}}{2}$$

Add $\frac{5}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{21}+5}{2}$$ $$x=\frac{5-\sqrt{21}}{2}$$

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Question

Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square