Solve \frac { 2 } { x ^ { 2 } - 1 } - \frac { 1 } { x + 1 } = \frac { 1 } { 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

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

Solve for x

$x=2\sqrt{2}-1\approx 1.828427125$
$x=-2\sqrt{2}-1\approx -3.828427125$

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 $-1,1$ since division by zero is not defined. Multiply both sides of the equation by $2\left(x-1\right)\left(x+1\right)$, the least common multiple of $x^{2}-1,x+1,2$.

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

Multiply $2$ and $2$ to get $4$.

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

To find the opposite of $2x-2$, find the opposite of each term.

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

Add $4$ and $2$ to get $6$.

$$6-2x=\left(x-1\right)\left(x+1\right)$$

Consider $\left(x-1\right)\left(x+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.

$$6-2x=x^{2}-1$$

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

$$6-2x-x^{2}=-1$$

Add $1$ to both sides.

$$6-2x-x^{2}+1=0$$

Add $6$ and $1$ to get $7$.

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

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.

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

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

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

Square $-2$.

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

Multiply $-4$ times $-1$.

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

Multiply $4$ times $7$.

$$x=\frac{-\left(-2\right)±\sqrt{4+28}}{2\left(-1\right)}$$

Add $4$ to $28$.

$$x=\frac{-\left(-2\right)±\sqrt{32}}{2\left(-1\right)}$$

Take the square root of $32$.

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

The opposite of $-2$ is $2$.

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

Multiply $2$ times $-1$.

$$x=\frac{2±4\sqrt{2}}{-2}$$

Now solve the equation $x=\frac{2±4\sqrt{2}}{-2}$ when $±$ is plus. Add $2$ to $4\sqrt{2}$.

$$x=\frac{4\sqrt{2}+2}{-2}$$

Divide $4\sqrt{2}+2$ by $-2$.

$$x=-2\sqrt{2}-1$$

Now solve the equation $x=\frac{2±4\sqrt{2}}{-2}$ when $±$ is minus. Subtract $4\sqrt{2}$ from $2$.

$$x=\frac{2-4\sqrt{2}}{-2}$$

Divide $2-4\sqrt{2}$ by $-2$.

$$x=2\sqrt{2}-1$$

The equation is now solved.

$$x=-2\sqrt{2}-1$$ $$x=2\sqrt{2}-1$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-1,1$ since division by zero is not defined. Multiply both sides of the equation by $2\left(x-1\right)\left(x+1\right)$, the least common multiple of $x^{2}-1,x+1,2$.

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

Multiply $2$ and $2$ to get $4$.

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

To find the opposite of $2x-2$, find the opposite of each term.

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

Add $4$ and $2$ to get $6$.

$$6-2x=\left(x-1\right)\left(x+1\right)$$

Consider $\left(x-1\right)\left(x+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.

$$6-2x=x^{2}-1$$

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

$$6-2x-x^{2}=-1$$

Subtract $6$ from both sides.

$$-2x-x^{2}=-1-6$$

Subtract $6$ from $-1$ to get $-7$.

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

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

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

Divide both sides by $-1$.

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

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

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

Divide $-2$ by $-1$.

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

Divide $-7$ by $-1$.

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

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

$$x^{2}+2x+1^{2}=7+1^{2}$$

Square $1$.

$$x^{2}+2x+1=7+1$$

Add $7$ to $1$.

$$x^{2}+2x+1=8$$

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

$$\left(x+1\right)^{2}=8$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{8}$$

Simplify.

$$x+1=2\sqrt{2}$$ $$x+1=-2\sqrt{2}$$

Subtract $1$ from both sides of the equation.

$$x=2\sqrt{2}-1$$ $$x=-2\sqrt{2}-1$$