Solve uation (2x + 1)/(x + 2) - (x - 2)/(x + 4) = 0 | 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

$$\frac { 2 x + 1 } { x + 2 } - \frac { x - 2 } { x + 4 } = 0$$

Solve for x

$x=-8$
$x=-1$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Variable $x$ cannot be equal to any of the values $-4,-2$ since division by zero is not defined. Multiply both sides of the equation by $\left(x+2\right)\left(x+4\right)$, the least common multiple of $x+2,x+4$.

$$\left(x+4\right)\left(2x+1\right)-\left(x+2\right)\left(x-2\right)=0$$

Use the distributive property to multiply $x+4$ by $2x+1$ and combine like terms.

$$2x^{2}+9x+4-\left(x+2\right)\left(x-2\right)=0$$

Consider $\left(x+2\right)\left(x-2\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $2$.

$$2x^{2}+9x+4-\left(x^{2}-4\right)=0$$

To find the opposite of $x^{2}-4$, find the opposite of each term.

$$2x^{2}+9x+4-x^{2}+4=0$$

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

$$x^{2}+9x+4+4=0$$

Add $4$ and $4$ to get $8$.

$$x^{2}+9x+8=0$$

To solve the equation, factor $x^{2}+9x+8$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=9$$ $$ab=8$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $8$.

$$1,8$$ $$2,4$$

Calculate the sum for each pair.

$$1+8=9$$ $$2+4=6$$

The solution is the pair that gives sum $9$.

$$a=1$$ $$b=8$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

$$\left(x+1\right)\left(x+8\right)$$

To find equation solutions, solve $x+1=0$ and $x+8=0$.

$$x=-1$$ $$x=-8$$

Steps Using Factoring By Grouping

Variable $x$ cannot be equal to any of the values $-4,-2$ since division by zero is not defined. Multiply both sides of the equation by $\left(x+2\right)\left(x+4\right)$, the least common multiple of $x+2,x+4$.

$$\left(x+4\right)\left(2x+1\right)-\left(x+2\right)\left(x-2\right)=0$$

Use the distributive property to multiply $x+4$ by $2x+1$ and combine like terms.

$$2x^{2}+9x+4-\left(x+2\right)\left(x-2\right)=0$$

Consider $\left(x+2\right)\left(x-2\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $2$.

$$2x^{2}+9x+4-\left(x^{2}-4\right)=0$$

To find the opposite of $x^{2}-4$, find the opposite of each term.

$$2x^{2}+9x+4-x^{2}+4=0$$

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

$$x^{2}+9x+4+4=0$$

Add $4$ and $4$ to get $8$.

$$x^{2}+9x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+8$. To find $a$ and $b$, set up a system to be solved.

$$a+b=9$$ $$ab=1\times 8=8$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $8$.

$$1,8$$ $$2,4$$

Calculate the sum for each pair.

$$1+8=9$$ $$2+4=6$$

The solution is the pair that gives sum $9$.

$$a=1$$ $$b=8$$

Rewrite $x^{2}+9x+8$ as $\left(x^{2}+x\right)+\left(8x+8\right)$.

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

Factor out $x$ in the first and $8$ in the second group.

$$x\left(x+1\right)+8\left(x+1\right)$$

Factor out common term $x+1$ by using distributive property.

$$\left(x+1\right)\left(x+8\right)$$

To find equation solutions, solve $x+1=0$ and $x+8=0$.

$$x=-1$$ $$x=-8$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-4,-2$ since division by zero is not defined. Multiply both sides of the equation by $\left(x+2\right)\left(x+4\right)$, the least common multiple of $x+2,x+4$.

$$\left(x+4\right)\left(2x+1\right)-\left(x+2\right)\left(x-2\right)=0$$

Use the distributive property to multiply $x+4$ by $2x+1$ and combine like terms.

$$2x^{2}+9x+4-\left(x+2\right)\left(x-2\right)=0$$

Consider $\left(x+2\right)\left(x-2\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $2$.

$$2x^{2}+9x+4-\left(x^{2}-4\right)=0$$

To find the opposite of $x^{2}-4$, find the opposite of each term.

$$2x^{2}+9x+4-x^{2}+4=0$$

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

$$x^{2}+9x+4+4=0$$

Add $4$ and $4$ to get $8$.

$$x^{2}+9x+8=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $9$ for $b$, and $8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-9±\sqrt{9^{2}-4\times 8}}{2}$$

Square $9$.

$$x=\frac{-9±\sqrt{81-4\times 8}}{2}$$

Multiply $-4$ times $8$.

$$x=\frac{-9±\sqrt{81-32}}{2}$$

Add $81$ to $-32$.

$$x=\frac{-9±\sqrt{49}}{2}$$

Take the square root of $49$.

$$x=\frac{-9±7}{2}$$

Now solve the equation $x=\frac{-9±7}{2}$ when $±$ is plus. Add $-9$ to $7$.

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

Divide $-2$ by $2$.

$$x=-1$$

Now solve the equation $x=\frac{-9±7}{2}$ when $±$ is minus. Subtract $7$ from $-9$.

$$x=-\frac{16}{2}$$

Divide $-16$ by $2$.

$$x=-8$$

The equation is now solved.

$$x=-1$$ $$x=-8$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-4,-2$ since division by zero is not defined. Multiply both sides of the equation by $\left(x+2\right)\left(x+4\right)$, the least common multiple of $x+2,x+4$.

$$\left(x+4\right)\left(2x+1\right)-\left(x+2\right)\left(x-2\right)=0$$

Use the distributive property to multiply $x+4$ by $2x+1$ and combine like terms.

$$2x^{2}+9x+4-\left(x+2\right)\left(x-2\right)=0$$

Consider $\left(x+2\right)\left(x-2\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $2$.

$$2x^{2}+9x+4-\left(x^{2}-4\right)=0$$

To find the opposite of $x^{2}-4$, find the opposite of each term.

$$2x^{2}+9x+4-x^{2}+4=0$$

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

$$x^{2}+9x+4+4=0$$

Add $4$ and $4$ to get $8$.

$$x^{2}+9x+8=0$$

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

$$x^{2}+9x=-8$$

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

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

Square $\frac{9}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+9x+\frac{81}{4}=-8+\frac{81}{4}$$

Add $-8$ to $\frac{81}{4}$.

$$x^{2}+9x+\frac{81}{4}=\frac{49}{4}$$

Factor $x^{2}+9x+\frac{81}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{9}{2}=\frac{7}{2}$$ $$x+\frac{9}{2}=-\frac{7}{2}$$

Subtract $\frac{9}{2}$ from both sides of the equation.

$$x=-1$$ $$x=-8$$