Solve 4/x - 8/(x + 3) = 1 | 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

$$\frac{4}{x}-\frac{8}{x+3}=1$$

Solve for x

$x = \frac{\sqrt{97} - 7}{2} \approx 1.424428901$
$x=\frac{-\sqrt{97}-7}{2}\approx -8.424428901$

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 $-3,0$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x+3\right)$, the least common multiple of $x,x+3$.

$$\left(x+3\right)\times 4-x\times 8=x\left(x+3\right)$$

Use the distributive property to multiply $x+3$ by $4$.

$$4x+12-x\times 8=x\left(x+3\right)$$

Use the distributive property to multiply $x$ by $x+3$.

$$4x+12-x\times 8=x^{2}+3x$$

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

$$4x+12-x\times 8-x^{2}=3x$$

Subtract $3x$ from both sides.

$$4x+12-x\times 8-x^{2}-3x=0$$

Combine $4x$ and $-3x$ to get $x$.

$$x+12-x\times 8-x^{2}=0$$

Multiply $-1$ and $8$ to get $-8$.

$$x+12-8x-x^{2}=0$$

Combine $x$ and $-8x$ to get $-7x$.

$$-7x+12-x^{2}=0$$

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.

$$-x^{2}-7x+12=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $-7$ for $b$, and $12$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Square $-7$.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-\left(-7\right)±\sqrt{49+4\times 12}}{2\left(-1\right)}$$

Multiply $4$ times $12$.

$$x=\frac{-\left(-7\right)±\sqrt{49+48}}{2\left(-1\right)}$$

Add $49$ to $48$.

$$x=\frac{-\left(-7\right)±\sqrt{97}}{2\left(-1\right)}$$

The opposite of $-7$ is $7$.

$$x=\frac{7±\sqrt{97}}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{7±\sqrt{97}}{-2}$$

Now solve the equation $x=\frac{7±\sqrt{97}}{-2}$ when $±$ is plus. Add $7$ to $\sqrt{97}$.

$$x=\frac{\sqrt{97}+7}{-2}$$

Divide $7+\sqrt{97}$ by $-2$.

$$x=\frac{-\sqrt{97}-7}{2}$$

Now solve the equation $x=\frac{7±\sqrt{97}}{-2}$ when $±$ is minus. Subtract $\sqrt{97}$ from $7$.

$$x=\frac{7-\sqrt{97}}{-2}$$

Divide $7-\sqrt{97}$ by $-2$.

$$x=\frac{\sqrt{97}-7}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{97}-7}{2}$$ $$x=\frac{\sqrt{97}-7}{2}$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-3,0$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x+3\right)$, the least common multiple of $x,x+3$.

$$\left(x+3\right)\times 4-x\times 8=x\left(x+3\right)$$

Use the distributive property to multiply $x+3$ by $4$.

$$4x+12-x\times 8=x\left(x+3\right)$$

Use the distributive property to multiply $x$ by $x+3$.

$$4x+12-x\times 8=x^{2}+3x$$

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

$$4x+12-x\times 8-x^{2}=3x$$

Subtract $3x$ from both sides.

$$4x+12-x\times 8-x^{2}-3x=0$$

Combine $4x$ and $-3x$ to get $x$.

$$x+12-x\times 8-x^{2}=0$$

Subtract $12$ from both sides. Anything subtracted from zero gives its negation.

$$x-x\times 8-x^{2}=-12$$

Multiply $-1$ and $8$ to get $-8$.

$$x-8x-x^{2}=-12$$

Combine $x$ and $-8x$ to get $-7x$.

$$-7x-x^{2}=-12$$

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 $x^{2}+bx=c$.

$$-x^{2}-7x=-12$$

Divide both sides by $-1$.

$$\frac{-x^{2}-7x}{-1}=-\frac{12}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\left(-\frac{7}{-1}\right)x=-\frac{12}{-1}$$

Divide $-7$ by $-1$.

$$x^{2}+7x=-\frac{12}{-1}$$

Divide $-12$ by $-1$.

$$x^{2}+7x=12$$

Divide $7$, the coefficient of the $x$ term, by $2$ to get $\frac{7}{2}$. Then add the square of $\frac{7}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=12+\left(\frac{7}{2}\right)^{2}$$

Square $\frac{7}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=12+\frac{49}{4}$$

Add $12$ to $\frac{49}{4}$.

$$x^{2}+7x+\frac{49}{4}=\frac{97}{4}$$

Factor $x^{2}+7x+\frac{49}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{97}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{97}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{\sqrt{97}}{2}$$ $$x+\frac{7}{2}=-\frac{\sqrt{97}}{2}$$

Subtract $\frac{7}{2}$ from both sides of the equation.

$$x=\frac{\sqrt{97}-7}{2}$$ $$x=\frac{-\sqrt{97}-7}{2}$$