If you've been paying close attention, it may have occurred to you that in this unit, every time we've completed a square, it's been a perfect square that we added to each side. This is important, because in one step of the solving process, we take the square root, and if both sides aren't perfect squares, then we could be headed for trouble.

Actually, though, it really isn't trouble - because numbers like 8 and 13 and 24 all have square roots, even if they aren't nice numbers.

So let's see what happens if we do a problem in which it doesn't come out to a nice perfect square.

Problem #1
Solve for x if x² + 4x + 2 = 0

Solution #1
To make a perfect square of the left side, we have to add 2 to both sides:

x² + 4x + 2 + 2 = 0 + 2
x² + 4x + 4 = 2
(x + 2)² = ()²
x + 2 = ±

Now if we subtract 2 from both sides, we have:

x = -2 ±

2

Problem #2
Solve for x if 2x² + 6x + 3 = 0

Solution #2
Multiply both sides by 2 to make the coefficient of x² a perfect square:

4x² + 12x + 6 = 0

To make this a perfect square, we need to add 3 to both sides:

4x² + 12x + 6 + 3 = 0 + 3
4x² + 12x + 9 = 3
(2x + 3)² = ()²
2x + 3 = ±
2x = -3  ±
x =

Problem #3
Find the solutions of x² - 6x + 2 = 0, to the nearest hundredth.

Solution #3
Since we're told to round to the nearest hundredth, we can't just leave our answer in radical form; we'll have to find the values using our calculators.

We need to add 7 to both sides of the equation to make a perfect square:

x² - 6x + 2 + 7 = 0 + 7
x² - 6x + 9 = 7
(x - 3)² = ()²
x - 3 = ±
x = 3 ±

x = 3 + or x = 3 -
x = 5.65 or x = .35

In the problems below, write your answers rounded to the nearest hundredth.

1.

x² + 5x + 3 = 0

2.

x² + 3x - 3 = 0

3.

x² - 4x - 2 = 0

4.

2x² + 2x - 1 = 0

5.

2x² + 4x -3 = 0