[SOLVED] Algebra2 Homework 3

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 (b) (x (c) (x (d) (x (e) (x (f) (x (g) (x (h) (x

)2=x2+2x+( )2 )2=x2−6x+( )2 )2=x2+3x+( )2 )2=x2−11x+( )2 )2=x2+21x+()2 )2=x2−25x+()2 )2=x2+37x+()2 )2=x2+abx+()2

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

a±c . That is, since you’re going to plug your answers back in to the original equations, then writing b

your answers with a common denominator will just make things more difficult. Here’s an example of what I’m looking for:

􏰉 1􏰊2
5 x+2 +10=0

􏰉 1􏰊2 5x+2 =−10

􏰉 1􏰊2
x + 2 = −2

􏰛􏰉 1􏰊2 √ x+2 =±−2

x + 21 = ± i √ 2 x = − 12 ± i √ 2

􏰋1√1√􏰌 x∈ −2+i 2,−2−i 2

Now we’ll check both answers simultaneously…

􏰉 1􏰊2 􏰉1 √ 1􏰊2
5 x+2 +10=5 −2±i 2+2 +10

􏰁 √􏰂2
=5 ±i 2 +10

= 5i2(2) + 10

= −5(2) + 10

= −10 + 10 = 0

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
(−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

x = − 12 ± i √ 2

√ x=−1±2i 2

√ x=−1±2i 2

2

which we would have plugged in to get a giant headache:

􏰙−1±2i√2 1􏰚2
5 2 +2 +10

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!)

and

22

(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

(e) 2 􏰍x − 41 􏰎2 − 38 = 0

(f) 3 􏰍x + 72 􏰎2 + 56 = 0 (g) a(x+h)2−k=0 (h) a(x−h)2+k=0

(assume0<k,0<a) (assume0<k,0<a)