Solve x/(x - 2) - (x - 2)/x = 5/2 | 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

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

Solve for x

$x=\frac{9-\sqrt{41}}{5}\approx 0.519375153$
$x = \frac{\sqrt{41} + 9}{5} \approx 3.080624847$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $0,2$ since division by zero is not defined. Multiply both sides of the equation by $2x\left(x-2\right)$, the least common multiple of $x-2,x,2$.

$$2xx-\left(2x-4\right)\left(x-2\right)=5x\left(x-2\right)$$

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

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

Use the distributive property to multiply $2x-4$ by $x-2$ and combine like terms.

$$2x^{2}-\left(2x^{2}-8x+8\right)=5x\left(x-2\right)$$

To find the opposite of $2x^{2}-8x+8$, find the opposite of each term.

$$2x^{2}-2x^{2}+8x-8=5x\left(x-2\right)$$

Combine $2x^{2}$ and $-2x^{2}$ to get $0$.

$$8x-8=5x\left(x-2\right)$$

Use the distributive property to multiply $5x$ by $x-2$.

$$8x-8=5x^{2}-10x$$

Subtract $5x^{2}$ from both sides.

$$8x-8-5x^{2}=-10x$$

Add $10x$ to both sides.

$$8x-8-5x^{2}+10x=0$$

Combine $8x$ and $10x$ to get $18x$.

$$18x-8-5x^{2}=0$$

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.

$$-5x^{2}+18x-8=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-5$ for $a$, $18$ for $b$, and $-8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-5\right)\left(-8\right)}}{2\left(-5\right)}$$

Square $18$.

$$x=\frac{-18±\sqrt{324-4\left(-5\right)\left(-8\right)}}{2\left(-5\right)}$$

Multiply $-4$ times $-5$.

$$x=\frac{-18±\sqrt{324+20\left(-8\right)}}{2\left(-5\right)}$$

Multiply $20$ times $-8$.

$$x=\frac{-18±\sqrt{324-160}}{2\left(-5\right)}$$

Add $324$ to $-160$.

$$x=\frac{-18±\sqrt{164}}{2\left(-5\right)}$$

Take the square root of $164$.

$$x=\frac{-18±2\sqrt{41}}{2\left(-5\right)}$$

Multiply $2$ times $-5$.

$$x=\frac{-18±2\sqrt{41}}{-10}$$

Now solve the equation $x=\frac{-18±2\sqrt{41}}{-10}$ when $±$ is plus. Add $-18$ to $2\sqrt{41}$.

$$x=\frac{2\sqrt{41}-18}{-10}$$

Divide $-18+2\sqrt{41}$ by $-10$.

$$x=\frac{9-\sqrt{41}}{5}$$

Now solve the equation $x=\frac{-18±2\sqrt{41}}{-10}$ when $±$ is minus. Subtract $2\sqrt{41}$ from $-18$.

$$x=\frac{-2\sqrt{41}-18}{-10}$$

Divide $-18-2\sqrt{41}$ by $-10$.

$$x=\frac{\sqrt{41}+9}{5}$$

The equation is now solved.

$$x=\frac{9-\sqrt{41}}{5}$$ $$x=\frac{\sqrt{41}+9}{5}$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $0,2$ since division by zero is not defined. Multiply both sides of the equation by $2x\left(x-2\right)$, the least common multiple of $x-2,x,2$.

$$2xx-\left(2x-4\right)\left(x-2\right)=5x\left(x-2\right)$$

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

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

Use the distributive property to multiply $2x-4$ by $x-2$ and combine like terms.

$$2x^{2}-\left(2x^{2}-8x+8\right)=5x\left(x-2\right)$$

To find the opposite of $2x^{2}-8x+8$, find the opposite of each term.

$$2x^{2}-2x^{2}+8x-8=5x\left(x-2\right)$$

Combine $2x^{2}$ and $-2x^{2}$ to get $0$.

$$8x-8=5x\left(x-2\right)$$

Use the distributive property to multiply $5x$ by $x-2$.

$$8x-8=5x^{2}-10x$$

Subtract $5x^{2}$ from both sides.

$$8x-8-5x^{2}=-10x$$

Add $10x$ to both sides.

$$8x-8-5x^{2}+10x=0$$

Combine $8x$ and $10x$ to get $18x$.

$$18x-8-5x^{2}=0$$

Add $8$ to both sides. Anything plus zero gives itself.

$$18x-5x^{2}=8$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-5x^{2}+18x=8$$

Divide both sides by $-5$.

$$\frac{-5x^{2}+18x}{-5}=\frac{8}{-5}$$

Dividing by $-5$ undoes the multiplication by $-5$.

$$x^{2}+\frac{18}{-5}x=\frac{8}{-5}$$

Divide $18$ by $-5$.

$$x^{2}-\frac{18}{5}x=\frac{8}{-5}$$

Divide $8$ by $-5$.

$$x^{2}-\frac{18}{5}x=-\frac{8}{5}$$

Divide $-\frac{18}{5}$, the coefficient of the $x$ term, by $2$ to get $-\frac{9}{5}$. Then add the square of $-\frac{9}{5}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{18}{5}x+\left(-\frac{9}{5}\right)^{2}=-\frac{8}{5}+\left(-\frac{9}{5}\right)^{2}$$

Square $-\frac{9}{5}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=-\frac{8}{5}+\frac{81}{25}$$

Add $-\frac{8}{5}$ to $\frac{81}{25}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{41}{25}$$

Factor $x^{2}-\frac{18}{5}x+\frac{81}{25}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{9}{5}\right)^{2}=\frac{41}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{5}\right)^{2}}=\sqrt{\frac{41}{25}}$$

Simplify.

$$x-\frac{9}{5}=\frac{\sqrt{41}}{5}$$ $$x-\frac{9}{5}=-\frac{\sqrt{41}}{5}$$

Add $\frac{9}{5}$ to both sides of the equation.

$$x=\frac{\sqrt{41}+9}{5}$$ $$x=\frac{9-\sqrt{41}}{5}$$