Solve sqrt{q+2}+1=sqrt{3q+7}

$$\sqrt { q + 2 } + 1 = \sqrt { 3 q + 7 }$$

Solve for q

$$q=-1$$

$$q=-2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{q+2}+1\right)^{2}=\left(\sqrt{3q+7}\right)^{2}$$

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

$$\left(\sqrt{q+2}\right)^{2}+2\sqrt{q+2}+1=\left(\sqrt{3q+7}\right)^{2}$$

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

$$q+2+2\sqrt{q+2}+1=\left(\sqrt{3q+7}\right)^{2}$$

Add 2 and 1 to get 3.

$$q+3+2\sqrt{q+2}=\left(\sqrt{3q+7}\right)^{2}$$

Calculate \sqrt{3q+7} to the power of 2 and get 3q+7.

$$q+3+2\sqrt{q+2}=3q+7$$

Subtract q+3 from both sides of the equation.

$$2\sqrt{q+2}=3q+7-\left(q+3\right)$$

To find the opposite of q+3, find the opposite of each term.

$$2\sqrt{q+2}=3q+7-q-3$$

Combine 3q and -q to get 2q.

$$2\sqrt{q+2}=2q+7-3$$

Subtract 3 from 7 to get 4.

$$2\sqrt{q+2}=2q+4$$

Square both sides of the equation.

$$\left(2\sqrt{q+2}\right)^{2}=\left(2q+4\right)^{2}$$

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

$$2^{2}\left(\sqrt{q+2}\right)^{2}=\left(2q+4\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{q+2}\right)^{2}=\left(2q+4\right)^{2}$$

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

$$4\left(q+2\right)=\left(2q+4\right)^{2}$$

Use the distributive property to multiply 4 by q+2.

$$4q+8=\left(2q+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2q+4\right)^{2}.

$$4q+8=4q^{2}+16q+16$$

Subtract 4q^{2} from both sides.

$$4q+8-4q^{2}=16q+16$$

Subtract 16q from both sides.

$$4q+8-4q^{2}-16q=16$$

Combine 4q and -16q to get -12q.

$$-12q+8-4q^{2}=16$$

Subtract 16 from both sides.

$$-12q+8-4q^{2}-16=0$$

Subtract 16 from 8 to get -8.

$$-12q-8-4q^{2}=0$$

Divide both sides by 4.

$$-3q-2-q^{2}=0$$

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

$$-q^{2}-3q-2=0$$

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

$$a+b=-3 ab=-\left(-2\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

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

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

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

Factor out q in the first and 2 in the second group.

$$q\left(-q-1\right)+2\left(-q-1\right)$$

Factor out common term -q-1 by using distributive property.

$$\left(-q-1\right)\left(q+2\right)$$

To find equation solutions, solve -q-1=0 and q+2=0.

$$q=-1 q=-2$$

Substitute -1 for q in the equation \sqrt{q+2}+1=\sqrt{3q+7}.

$$\sqrt{-1+2}+1=\sqrt{3\left(-1\right)+7}$$

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

$$2=2$$

Substitute -2 for q in the equation \sqrt{q+2}+1=\sqrt{3q+7}.

$$\sqrt{-2+2}+1=\sqrt{3\left(-2\right)+7}$$

Simplify. The value q=-2 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{q+2}+1=\sqrt{3q+7}.

$$q=-1 q=-2$$