Solve DATE LCD 4/x + 8/(x + 2) = 4 | 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

$$\frac { 4 } { x } + \frac { 8 } { x + 2 } = 4$$

Solve for x

$x=-1$
$x=2$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

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

$$\left(x+2\right)\times 4+x\times 8=4x\left(x+2\right)$$

Use the distributive property to multiply $x+2$ by $4$.

$$4x+8+x\times 8=4x\left(x+2\right)$$

Combine $4x$ and $x\times 8$ to get $12x$.

$$12x+8=4x\left(x+2\right)$$

Use the distributive property to multiply $4x$ by $x+2$.

$$12x+8=4x^{2}+8x$$

Subtract $4x^{2}$ from both sides.

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

Subtract $8x$ from both sides.

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

Combine $12x$ and $-8x$ to get $4x$.

$$4x+8-4x^{2}=0$$

Divide both sides by $4$.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx+2$. To find $a$ and $b$, set up a system to be solved.

$$a+b=1$$ $$ab=-2=-2$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=2$$ $$b=-1$$

Rewrite $-x^{2}+x+2$ as $\left(-x^{2}+2x\right)+\left(-x+2\right)$.

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

Factor out $-x$ in the first and $-1$ in the second group.

$$-x\left(x-2\right)-\left(x-2\right)$$

Factor out common term $x-2$ by using distributive property.

$$\left(x-2\right)\left(-x-1\right)$$

To find equation solutions, solve $x-2=0$ and $-x-1=0$.

$$x=2$$ $$x=-1$$

Steps Using the Quadratic Formula

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

$$\left(x+2\right)\times 4+x\times 8=4x\left(x+2\right)$$

Use the distributive property to multiply $x+2$ by $4$.

$$4x+8+x\times 8=4x\left(x+2\right)$$

Combine $4x$ and $x\times 8$ to get $12x$.

$$12x+8=4x\left(x+2\right)$$

Use the distributive property to multiply $4x$ by $x+2$.

$$12x+8=4x^{2}+8x$$

Subtract $4x^{2}$ from both sides.

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

Subtract $8x$ from both sides.

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

Combine $12x$ and $-8x$ to get $4x$.

$$4x+8-4x^{2}=0$$

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.

$$-4x^{2}+4x+8=0$$

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

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

Square $4$.

$$x=\frac{-4±\sqrt{16-4\left(-4\right)\times 8}}{2\left(-4\right)}$$

Multiply $-4$ times $-4$.

$$x=\frac{-4±\sqrt{16+16\times 8}}{2\left(-4\right)}$$

Multiply $16$ times $8$.

$$x=\frac{-4±\sqrt{16+128}}{2\left(-4\right)}$$

Add $16$ to $128$.

$$x=\frac{-4±\sqrt{144}}{2\left(-4\right)}$$

Take the square root of $144$.

$$x=\frac{-4±12}{2\left(-4\right)}$$

Multiply $2$ times $-4$.

$$x=\frac{-4±12}{-8}$$

Now solve the equation $x=\frac{-4±12}{-8}$ when $±$ is plus. Add $-4$ to $12$.

$$x=\frac{8}{-8}$$

Divide $8$ by $-8$.

$$x=-1$$

Now solve the equation $x=\frac{-4±12}{-8}$ when $±$ is minus. Subtract $12$ from $-4$.

$$x=-\frac{16}{-8}$$

Divide $-16$ by $-8$.

$$x=2$$

The equation is now solved.

$$x=-1$$ $$x=2$$

Steps for Completing the Square

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

$$\left(x+2\right)\times 4+x\times 8=4x\left(x+2\right)$$

Use the distributive property to multiply $x+2$ by $4$.

$$4x+8+x\times 8=4x\left(x+2\right)$$

Combine $4x$ and $x\times 8$ to get $12x$.

$$12x+8=4x\left(x+2\right)$$

Use the distributive property to multiply $4x$ by $x+2$.

$$12x+8=4x^{2}+8x$$

Subtract $4x^{2}$ from both sides.

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

Subtract $8x$ from both sides.

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

Combine $12x$ and $-8x$ to get $4x$.

$$4x+8-4x^{2}=0$$

Subtract $8$ from both sides. Anything subtracted from zero gives its negation.

$$4x-4x^{2}=-8$$

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

$$-4x^{2}+4x=-8$$

Divide both sides by $-4$.

$$\frac{-4x^{2}+4x}{-4}=-\frac{8}{-4}$$

Dividing by $-4$ undoes the multiplication by $-4$.

$$x^{2}+\frac{4}{-4}x=-\frac{8}{-4}$$

Divide $4$ by $-4$.

$$x^{2}-x=-\frac{8}{-4}$$

Divide $-8$ by $-4$.

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

Divide $-1$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{2}$. Then add the square of $-\frac{1}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $-\frac{1}{2}$ by squaring both the numerator and the denominator of the fraction.

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

Add $2$ to $\frac{1}{4}$.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

Factor $x^{2}-x+\frac{1}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{3}{2}$$ $$x-\frac{1}{2}=-\frac{3}{2}$$

Add $\frac{1}{2}$ to both sides of the equation.

$$x=2$$ $$x=-1$$