Solve q^2+4q+4-2q^2=8

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

Solve for q

$$q=2$$

Steps Using Factoring By Grouping

Combine q^{2} and -2q^{2} to get -q^{2}.

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

Subtract 8 from both sides.

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

Subtract 8 from 4 to get -4.

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

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

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

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 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

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

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

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

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

Factor out common term q-2 by using distributive property.

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

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

$$q=2 q=2$$

Steps Using the Quadratic Formula

Combine q^{2} and -2q^{2} to get -q^{2}.

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

Subtract 8 from both sides.

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

Subtract 8 from 4 to get -4.

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

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

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

Square 4.

$$q=\frac{-4±\sqrt{16-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$q=\frac{-4±\sqrt{16+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$q=\frac{-4±\sqrt{16-16}}{2\left(-1\right)}$$

Add 16 to -16.

$$q=\frac{-4±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$q=-\frac{4}{2\left(-1\right)}$$

Multiply 2 times -1.

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

Divide -4 by -2.

$$q=2$$

Steps for Completing the Square

Combine q^{2} and -2q^{2} to get -q^{2}.

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

Subtract 4 from both sides.

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

Subtract 4 from 8 to get 4.

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

Divide both sides by -1.

$$\frac{-q^{2}+4q}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$q^{2}+\frac{4}{-1}q=\frac{4}{-1}$$

Divide 4 by -1.

$$q^{2}-4q=\frac{4}{-1}$$

Divide 4 by -1.

$$q^{2}-4q=-4$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add -4 to 4.

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

Factor q^{2}-4q+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q-2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$q-2=0 q-2=0$$

Add 2 to both sides of the equation.

$$q=2 q=2$$

The equation is now solved. Solutions are the same.

$$q=2$$