Solve sqrt{8-p}-sqrt{p+5}=1

$$\sqrt { 8 - p } - \sqrt { p + 5 } = 1$$

Solve for p

$$p=-1$$

Solution Steps

Subtract -\sqrt{p+5} from both sides of the equation.

$$\sqrt{8-p}=1+\sqrt{p+5}$$

Square both sides of the equation.

$$\left(\sqrt{8-p}\right)^{2}=\left(1+\sqrt{p+5}\right)^{2}$$

Calculate \sqrt{8-p} to the power of 2 and get 8-p.

$$8-p=\left(1+\sqrt{p+5}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{p+5}\right)^{2}.

$$8-p=1+2\sqrt{p+5}+\left(\sqrt{p+5}\right)^{2}$$

Calculate \sqrt{p+5} to the power of 2 and get p+5.

$$8-p=1+2\sqrt{p+5}+p+5$$

Add 1 and 5 to get 6.

$$8-p=6+2\sqrt{p+5}+p$$

Subtract 6+p from both sides of the equation.

$$8-p-\left(6+p\right)=2\sqrt{p+5}$$

To find the opposite of 6+p, find the opposite of each term.

$$8-p-6-p=2\sqrt{p+5}$$

Subtract 6 from 8 to get 2.

$$2-p-p=2\sqrt{p+5}$$

Combine -p and -p to get -2p.

$$2-2p=2\sqrt{p+5}$$

Square both sides of the equation.

$$\left(2-2p\right)^{2}=\left(2\sqrt{p+5}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-2p\right)^{2}.

$$4-8p+4p^{2}=\left(2\sqrt{p+5}\right)^{2}$$

Expand \left(2\sqrt{p+5}\right)^{2}.

$$4-8p+4p^{2}=2^{2}\left(\sqrt{p+5}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4-8p+4p^{2}=4\left(\sqrt{p+5}\right)^{2}$$

Calculate \sqrt{p+5} to the power of 2 and get p+5.

$$4-8p+4p^{2}=4\left(p+5\right)$$

Use the distributive property to multiply 4 by p+5.

$$4-8p+4p^{2}=4p+20$$

Subtract 4p from both sides.

$$4-8p+4p^{2}-4p=20$$

Combine -8p and -4p to get -12p.

$$4-12p+4p^{2}=20$$

Subtract 20 from both sides.

$$4-12p+4p^{2}-20=0$$

Subtract 20 from 4 to get -16.

$$-16-12p+4p^{2}=0$$

Divide both sides by 4.

$$-4-3p+p^{2}=0$$

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

$$p^{2}-3p-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp-4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-4\right)=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

Rewrite p^{2}-3p-4 as \left(p^{2}-4p\right)+\left(p-4\right).

$$\left(p^{2}-4p\right)+\left(p-4\right)$$

Factor out p in p^{2}-4p.

$$p\left(p-4\right)+p-4$$

Factor out common term p-4 by using distributive property.

$$\left(p-4\right)\left(p+1\right)$$

To find equation solutions, solve p-4=0 and p+1=0.

$$p=4 p=-1$$

Substitute 4 for p in the equation \sqrt{8-p}-\sqrt{p+5}=1.

$$\sqrt{8-4}-\sqrt{4+5}=1$$

Simplify. The value p=4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Substitute -1 for p in the equation \sqrt{8-p}-\sqrt{p+5}=1.

$$\sqrt{8-\left(-1\right)}-\sqrt{-1+5}=1$$

Simplify. The value p=-1 satisfies the equation.

$$1=1$$

Equation \sqrt{8-p}=\sqrt{p+5}+1 has a unique solution.

$$p=-1$$