Advanced Algebra and Trig question...?

Question

I actually have a few.

Solve by completing the square. Give both real and imaginary roots. Be sure not to lose or gain roots.

22) (2x+1)(4x-3)= 3(4x-3)^2
26) (r)/(r-1)-(r)/(r+1)= (2)/(r^2-1)
28) (x+2)/(x^2-x-6)= 3-(4)/(x-3)
30) *rad*2x+5*end rad*=x+1

I appreciate any help that anyone can give me. I know it' won't be easy. :S Why else would I be asking?
KH

Answer 1

22) (2x+1)(4x-3)= 3(4x-3)^2
8x^2 - 2x - 3 = 48x^2 - 72x + 27
40x^2 - 70x + 30 = 0
4x^2 - 7x + 3 = 0

I'd factor....
(4x - 3)(x - 1) = 0
x = 3/4, x = 1

But you say you want the square completed:

4x^2 - 7x + 3 = 0
x^2 - (7/4)x + 3/4 = 0
x^2 - (7/4)x = -3/4
x^2 - (7/4)x + 49/64 = -3/4 + 49/64
(x - 7/8)^2 = 1/64
(x - 7/8)^2 = (1/8)^2
x - 7/8 = +/- 1/8
x = 7/8 +/- 1/8

x = 7/8 + 1/8 = 1
x = 7/8 - 1/8 = 3/4

26) (r)/(r-1)-(r)/(r+1)= (2)/(r^2-1)

Multiply by (r+1)(r-1) to eliminate all denominators

r(r+1) - r(r - 1) = 2
r^2 + r - r^2 + r = 2
2r = 2
r = 1

We get r=1, but r=1 and r=-1 are ruled out by the original equation, which has both (r-1) and (r+1) in the denominators. So there are no solutions to this one.

28) (x+2)/(x^2-x-6)= 3-(4)/(x-3)

Multiply by (x-3)(x+2) [aka x^2 - x - 6] to eliminate denominators

x + 2 = 3(x^2 - x - 6) - 4(x + 2)
x + 2 = 3x^2 - 3x - 18 - 4x - 8
3x^2 - 8x - 28 = 0
(3x - 14)(x + 2) = 0

x = 14/3, x = -2

Completing the square:
3x^2 - 8x - 28 = 0
x^2 - (8/3)x = 28/3
x^2 - (8/3)x + 64/36 = 28/3 + 64/36
(x - 8/6)^2 = 400/36
(x - 8/6)^2 = (20/6)^2
x - 8/6 = +/- 20/6
x = 8/6 +/- 20/6

x = 8/6 + 20/6 = 28/6 = 14/3
x = 8/6 - 20/6 = -12/6 = -2

30) sqrt(2x + 5) = x + 1

Square both sides.

2x + 5 = x^2 + 2x + 1
x^2 = 4

Completing the square:

x^2 = 2^2
x = +/-2

x = 2
x = -2

Checking the solutions:

sqrt(2x + 5) =? x + 1
sqrt(2*2 + 5) =? 2 + 1
sqrt(9) = 3
check, x=2 works

sqrt(2x + 5) =? x + 1
sqrt(2*(-2) + 5) =? (-2) + 1
sqrt(1) =? -1
no, x=-2 doesn't work.

Answer 2

The first thing to do with all of these is to do the algebra necessary to get them into the form ax^2 + bx + c = 0. They can then be solved by any of the standard methods. I'll do some of the last one: squaring, get (2x + 5) = x^2 + 2x + 1; x^2 = 4, so x = +/- 2. Note that any time you have a square root in the original, you have to test the apparent roots to see if all are valid; in this case, they are.

Answer 3

It is your homework not ours. Ask questions more specific after you had done some work. If you are staying in class should have some idea.Your textbook is the better helper if you want learn mathematics rather show your homework done to your teacher.
I give you some guidelines:
1. Simplify equations and equals to zero.
2.Factorize using x^2 coefficient likes common factor.
3.Complete the square.
4.Solve for x.
If you have radicals , eliminate them first.