Solve sqrt{q-2}+3=sqrt{4q+1}

$$\sqrt{ q-2 } +3 = \sqrt{ 4q+1 }$$

Solve for q

$$q=6$$

$$q=2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{q-2}+3\right)^{2}=\left(\sqrt{4q+1}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{q-2}+3\right)^{2}.

$$\left(\sqrt{q-2}\right)^{2}+6\sqrt{q-2}+9=\left(\sqrt{4q+1}\right)^{2}$$

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

$$q-2+6\sqrt{q-2}+9=\left(\sqrt{4q+1}\right)^{2}$$

Add -2 and 9 to get 7.

$$q+7+6\sqrt{q-2}=\left(\sqrt{4q+1}\right)^{2}$$

Calculate \sqrt{4q+1} to the power of 2 and get 4q+1.

$$q+7+6\sqrt{q-2}=4q+1$$

Subtract q+7 from both sides of the equation.

$$6\sqrt{q-2}=4q+1-\left(q+7\right)$$

To find the opposite of q+7, find the opposite of each term.

$$6\sqrt{q-2}=4q+1-q-7$$

Combine 4q and -q to get 3q.

$$6\sqrt{q-2}=3q+1-7$$

Subtract 7 from 1 to get -6.

$$6\sqrt{q-2}=3q-6$$

Square both sides of the equation.

$$\left(6\sqrt{q-2}\right)^{2}=\left(3q-6\right)^{2}$$

Expand \left(6\sqrt{q-2}\right)^{2}.

$$6^{2}\left(\sqrt{q-2}\right)^{2}=\left(3q-6\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$36\left(\sqrt{q-2}\right)^{2}=\left(3q-6\right)^{2}$$

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

$$36\left(q-2\right)=\left(3q-6\right)^{2}$$

Use the distributive property to multiply 36 by q-2.

$$36q-72=\left(3q-6\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3q-6\right)^{2}.

$$36q-72=9q^{2}-36q+36$$

Subtract 9q^{2} from both sides.

$$36q-72-9q^{2}=-36q+36$$

Add 36q to both sides.

$$36q-72-9q^{2}+36q=36$$

Combine 36q and 36q to get 72q.

$$72q-72-9q^{2}=36$$

Subtract 36 from both sides.

$$72q-72-9q^{2}-36=0$$

Subtract 36 from -72 to get -108.

$$72q-108-9q^{2}=0$$

Divide both sides by 9.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -q^{2}+aq+bq-12. To find a and b, set up a system to be solved.

$$a+b=8 ab=-\left(-12\right)=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=6 b=2$$

Rewrite -q^{2}+8q-12 as \left(-q^{2}+6q\right)+\left(2q-12\right).

$$\left(-q^{2}+6q\right)+\left(2q-12\right)$$

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

$$-q\left(q-6\right)+2\left(q-6\right)$$

Factor out common term q-6 by using distributive property.

$$\left(q-6\right)\left(-q+2\right)$$

To find equation solutions, solve q-6=0 and -q+2=0.

$$q=6 q=2$$

Substitute 6 for q in the equation \sqrt{q-2}+3=\sqrt{4q+1}.

$$\sqrt{6-2}+3=\sqrt{4\times 6+1}$$

Simplify. The value q=6 satisfies the equation.

$$5=5$$

Substitute 2 for q in the equation \sqrt{q-2}+3=\sqrt{4q+1}.

$$\sqrt{2-2}+3=\sqrt{4\times 2+1}$$

Simplify. The value q=2 satisfies the equation.

$$3=3$$

List all solutions of \sqrt{q-2}+3=\sqrt{4q+1}.

$$q=6 q=2$$