Solve 1/(4 - p) - 1/(2 + p) = 1/4 | 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

$$\frac{1}{4-p}-\frac{1}{2+p}=\frac{1}{4}$$

Solve for p

$p=-8$
$p=2$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$-8-4p-\left(4p-16\right)=\left(p-4\right)\left(p+2\right)$$

To find the opposite of $4p-16$, find the opposite of each term.

$$-8-4p-4p+16=\left(p-4\right)\left(p+2\right)$$

Combine $-4p$ and $-4p$ to get $-8p$.

$$-8-8p+16=\left(p-4\right)\left(p+2\right)$$

Add $-8$ and $16$ to get $8$.

$$8-8p=\left(p-4\right)\left(p+2\right)$$

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

$$8-8p=p^{2}-2p-8$$

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

$$8-8p-p^{2}=-2p-8$$

Add $2p$ to both sides.

$$8-8p-p^{2}+2p=-8$$

Combine $-8p$ and $2p$ to get $-6p$.

$$8-6p-p^{2}=-8$$

Add $8$ to both sides.

$$8-6p-p^{2}+8=0$$

Add $8$ and $8$ to get $16$.

$$16-6p-p^{2}=0$$

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.

$$-p^{2}-6p+16=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $-6$ for $b$, and $16$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$p=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

Square $-6$.

$$p=\frac{-\left(-6\right)±\sqrt{36-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$p=\frac{-\left(-6\right)±\sqrt{36+4\times 16}}{2\left(-1\right)}$$

Multiply $4$ times $16$.

$$p=\frac{-\left(-6\right)±\sqrt{36+64}}{2\left(-1\right)}$$

Add $36$ to $64$.

$$p=\frac{-\left(-6\right)±\sqrt{100}}{2\left(-1\right)}$$

Take the square root of $100$.

$$p=\frac{-\left(-6\right)±10}{2\left(-1\right)}$$

The opposite of $-6$ is $6$.

$$p=\frac{6±10}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$p=\frac{6±10}{-2}$$

Now solve the equation $p=\frac{6±10}{-2}$ when $±$ is plus. Add $6$ to $10$.

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

Divide $16$ by $-2$.

$$p=-8$$

Now solve the equation $p=\frac{6±10}{-2}$ when $±$ is minus. Subtract $10$ from $6$.

$$p=-\frac{4}{-2}$$

Divide $-4$ by $-2$.

$$p=2$$

The equation is now solved.

$$p=-8$$ $$p=2$$

Steps for Completing the Square

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

$$-8-4p-\left(4p-16\right)=\left(p-4\right)\left(p+2\right)$$

To find the opposite of $4p-16$, find the opposite of each term.

$$-8-4p-4p+16=\left(p-4\right)\left(p+2\right)$$

Combine $-4p$ and $-4p$ to get $-8p$.

$$-8-8p+16=\left(p-4\right)\left(p+2\right)$$

Add $-8$ and $16$ to get $8$.

$$8-8p=\left(p-4\right)\left(p+2\right)$$

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

$$8-8p=p^{2}-2p-8$$

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

$$8-8p-p^{2}=-2p-8$$

Add $2p$ to both sides.

$$8-8p-p^{2}+2p=-8$$

Combine $-8p$ and $2p$ to get $-6p$.

$$8-6p-p^{2}=-8$$

Subtract $8$ from both sides.

$$-6p-p^{2}=-8-8$$

Subtract $8$ from $-8$ to get $-16$.

$$-6p-p^{2}=-16$$

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

$$-p^{2}-6p=-16$$

Divide both sides by $-1$.

$$\frac{-p^{2}-6p}{-1}=-\frac{16}{-1}$$

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

$$p^{2}+\left(-\frac{6}{-1}\right)p=-\frac{16}{-1}$$

Divide $-6$ by $-1$.

$$p^{2}+6p=-\frac{16}{-1}$$

Divide $-16$ by $-1$.

$$p^{2}+6p=16$$

Divide $6$, the coefficient of the $x$ term, by $2$ to get $3$. Then add the square of $3$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+6p+3^{2}=16+3^{2}$$

Square $3$.

$$p^{2}+6p+9=16+9$$

Add $16$ to $9$.

$$p^{2}+6p+9=25$$

Factor $p^{2}+6p+9$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(p+3\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+3\right)^{2}}=\sqrt{25}$$

Simplify.

$$p+3=5$$ $$p+3=-5$$

Subtract $3$ from both sides of the equation.

$$p=2$$ $$p=-8$$