Solve x - (1 / (- x)) = 5 | 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

$$x-(1/-x)=5$$

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

Factor $-x$.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply $x$ times $\frac{-x}{-x}$.

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

Since $\frac{x\left(-1\right)x}{-x}$ and $\frac{1}{-x}$ have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in $x\left(-1\right)x-1$.

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

Subtract $5$ from both sides.

$$\frac{-x^{2}-1}{-x}-5=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $5$ times $\frac{-x}{-x}$.

$$\frac{-x^{2}-1}{-x}-\frac{5\left(-1\right)x}{-x}=0$$

Since $\frac{-x^{2}-1}{-x}$ and $\frac{5\left(-1\right)x}{-x}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{-x^{2}-1-5\left(-1\right)x}{-x}=0$$

Do the multiplications in $-x^{2}-1-5\left(-1\right)x$.

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

Factor the expressions that are not already factored in $\frac{-x^{2}-1+5x}{-x}$.

$$\frac{-\left(x-\left(-\frac{1}{2}\sqrt{21}+\frac{5}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{21}+\frac{5}{2}\right)\right)}{-x}=0$$

Cancel out $-1$ in both numerator and denominator.

$$\frac{\left(x-\left(-\frac{1}{2}\sqrt{21}+\frac{5}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{21}+\frac{5}{2}\right)\right)}{x}=0$$

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

$$\left(x-\left(-\frac{1}{2}\sqrt{21}+\frac{5}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{21}+\frac{5}{2}\right)\right)=0$$

To find the opposite of $-\frac{1}{2}\sqrt{21}+\frac{5}{2}$, find the opposite of each term.

$$\left(x+\frac{1}{2}\sqrt{21}-\frac{5}{2}\right)\left(x-\left(\frac{1}{2}\sqrt{21}+\frac{5}{2}\right)\right)=0$$

To find the opposite of $\frac{1}{2}\sqrt{21}+\frac{5}{2}$, find the opposite of each term.

$$\left(x+\frac{1}{2}\sqrt{21}-\frac{5}{2}\right)\left(x-\frac{1}{2}\sqrt{21}-\frac{5}{2}\right)=0$$

Use the distributive property to multiply $x+\frac{1}{2}\sqrt{21}-\frac{5}{2}$ by $x-\frac{1}{2}\sqrt{21}-\frac{5}{2}$ and combine like terms.

$$x^{2}-5x-\frac{1}{4}\left(\sqrt{21}\right)^{2}+\frac{25}{4}=0$$

The square of $\sqrt{21}$ is $21$.

$$x^{2}-5x-\frac{1}{4}\times 21+\frac{25}{4}=0$$

Multiply $-\frac{1}{4}$ and $21$ to get $-\frac{21}{4}$.

$$x^{2}-5x-\frac{21}{4}+\frac{25}{4}=0$$

Add $-\frac{21}{4}$ and $\frac{25}{4}$ to get $1$.

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

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

Factor $-x$.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply $x$ times $\frac{-x}{-x}$.

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

Since $\frac{x\left(-1\right)x}{-x}$ and $\frac{1}{-x}$ have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in $x\left(-1\right)x-1$.

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

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

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

To find the opposite of $-x^{2}-1$, find the opposite of each term.

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

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.

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