Problem 1. Fill in the missing information in each equation.
For example, given (x )2 = x2 −4x +( )2, you’ll need to complete it with the −2 on the left hand side, and with a 2 on the right giving you (x − 2)2 = x2 − 4x + (2)2. (Notice that it’s not necessary to write (−2)2 since it’s gonna be positive anyway. I mean, (−2)2 = (2)2 = 4, so why waste your time writing an unnecessary symbol?)
(a) (x )2 = x2+2x +( )2
(b) (x )2 = x2−6x +( )2
(c) (x )2 = x2+3x +( )2
(d) (x )2 = x2−11x +( )2
(e) (x )
(f) (x
(g) (x
(h) (x
Problem 2. Solve for x. Write your answers in a solution set. Check your answers. You’ll notice
that keeping your answers in the form ab ± cb will make it easier to check than if you’d written it like
. That is, since you’re going to plug your answers back in to the original equations, then writing your answers with a common denominator will just make things more difficult. Here’s an example of what I’m looking for:
Now we’ll check both answers simultaneously...
Note that the only reason that we can check both answers simultaneously is that both (a)2 = a2 and (−a)2 = a2. And as we just saw, this is even true for complex numbers since
√ √ √ √ √
(i 2)2 = i 2∗ i 2 = i2 2 2 = (−1)2 = −2
and
√ √ √ √ √ √ √
(−i 2)2 = −i 2∗−i 2 = (−1)i 2∗(−1)i 2 = (−1)2i2 2 2 = (1)(−1)2 = −2
So as long as we’re on the same page and understand that when we write (±a)2 = a2 we mean that both (a)2 = a2 and that (−a)2 = a2, then we can simplify our work by killing two birds with one stone.
Finally, notice how much harder it would have been to check if I’d written my answers with a common denominator. I would have had
which we would have plugged in to get a giant headache:
It’s much easier to keep these guys split up instead of putting them all over a common denominator.
(It’s usually easier to not rationalize denominators too!)
(a) (x −2)2−9 = 0
(b) (x +3)2−3 = 0
(c) (x −3)2+3 = 0
(d) 2(x +1)2−8 = 0
(g) a(x + h)2− k = 0
(assume 0 < k,0 < a)
(h) a(x − h)2+ k = 0
(assume 0 < k,0 < a)
