Algebra Honors ( Period 4)

Perfect Squares 8-9 
SOLVING QUADRATICS USING SQUARE ROOTING

We’ve been solving all quadratics with GRAPHING or the ZERO PRODUCTS PROPERTY after we get our happy face on and factor.

There are several other ways to solve 2nd degree equations:
Square rooting both sides – Chapter 8-9 Example 4 on p. 525 with Chapter 10-2
Completing the square and then square rooting both sides - Chapter 9-4
The Quadratic Formula – Chapter 9-5
(FYI: we’ll be skipping Chapter 9-6 and 9-7)

Think about a simple equation like x² = 36.
We could get our happy face on and factor:
x² – 36 = 0   (x + 6)(x-6) = 0
x = -6 and x = 6

However, there’s a much easier way to solve this quadratic: simply take the square root of each side!
x² = 36
√ x² = ±√36

WHY ±?
You need both of the roots (solutions, zeros, x-intercepts).
Why does this work so well?
There are only TWO TERMS: the quadratic term and the constant term…the linear x term is missing.

Consider a different equation such as x² + 5x = 36
Notice that I can no longer just take the square root of each side.
I would have to use the Zero Products Property:
x² + 5x -36 = 0
(x + 9)(x – 4) = 0
x = -9 and x = 4
If you have a BINOMIAL SQUARED, you can also use this approach:

(5x – 7)² = 16
√(5x – 7)² = ±√16
5x – 7 = ±4

5x – 7 = 4
and
5x – 7 = -4
solve
x = 11/5 and x = 3/5

If you have a TRINOMIAL SQUARE, factor it to a BINOMIAL SQUARED, and then use this approach:

25x² + 40x +16 = 49
(5x + 4)² = 49
√(5x + 4)² = ±√49
5x + 4 = ±7
5x + 4 = -7 and 5x + 4 = 7
solve both
x = -11/5 and x = 3/5

This works because you still have only 2 terms, one on each side, that you can take the square root of.

WHAT HAPPENS IF THE CONSTANT IS NOT A PERFECT SQUARE?
You’ll get an IRRATIONAL ANSWER.

There are 2 ways to give that answer.
1. Use the square root button on a calculator and ROUND (usually to the tenths or hundredths)
2. SIMPLIFY the square root (SEE ABOVE CHAPTER 10-2)