Solve q+2=2sqrt{4q-7}

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

Solve for q

$$q=8$$

$$q=4$$

Solution Steps

Square both sides of the equation.

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

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

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

Expand \left(2\sqrt{4q-7}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Calculate \sqrt{4q-7} to the power of 2 and get 4q-7.

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

Use the distributive property to multiply 4 by 4q-7.

$$q^{2}+4q+4=16q-28$$

Subtract 16q from both sides.

$$q^{2}+4q+4-16q=-28$$

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

$$q^{2}-12q+4=-28$$

Add 28 to both sides.

$$q^{2}-12q+4+28=0$$

Add 4 and 28 to get 32.

$$q^{2}-12q+32=0$$

To solve the equation, factor q^{2}-12q+32 using formula q^{2}+\left(a+b\right)q+ab=\left(q+a\right)\left(q+b\right). To find a and b, set up a system to be solved.

$$a+b=-12 ab=32$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 32.

$$-1,-32 -2,-16 -4,-8$$

Calculate the sum for each pair.

$$-1-32=-33 -2-16=-18 -4-8=-12$$

The solution is the pair that gives sum -12.

$$a=-8 b=-4$$

Rewrite factored expression \left(q+a\right)\left(q+b\right) using the obtained values.

$$\left(q-8\right)\left(q-4\right)$$

To find equation solutions, solve q-8=0 and q-4=0.

$$q=8 q=4$$

Substitute 8 for q in the equation q+2=2\sqrt{4q-7}.

$$8+2=2\sqrt{4\times 8-7}$$

Simplify. The value q=8 satisfies the equation.

$$10=10$$

Substitute 4 for q in the equation q+2=2\sqrt{4q-7}.

$$4+2=2\sqrt{4\times 4-7}$$

Simplify. The value q=4 satisfies the equation.

$$6=6$$

List all solutions of q+2=2\sqrt{4q-7}.

$$q=8 q=4$$