Salutations,

in class today, for all you naughty kids that skipped (ha, no I'm just kidding), we worked on some problems that Mr. K had put up on the board. We mainly focused on algebraic equations that deal with extraneous solutions and we also had breifly discussed absolute. The four questions that the class worked on were:

x= √x-2 +4

x-4= √x-2

(x-4)(x-4)= (√x-2)(√x-2)

x²-8x+16= x-2

x²-9x+18= 0

(x-6)(x-3)

x=6 or x=3

*REMINDER: Always begin solving your algebraic equation by isolating the variable!

To check which answer is correct, we substitute the two solutions we got.

6= √6-2 +4

6= √4 +4

6= 2+4

6= 6

Therefore 6 is the correct answer.

3= √3-2 +4

3= √1 +4

3=5

The number 3 does not work out so we reject it. It is an extraneous solution.

√4x+5 - √2x-1= 2

√4x+5= 2+√2x-1

(√4x+5)(√4x+5)= (2+√2x-1)<2√2x-1)

4x+5= 4+4√2x-1 +2x-1

2x+2= 4(2x+1)

x+1= 2√2x-1

(x+1)(x+1)= 4(2x-1)

x²+2x+1= 8x-4

x²-6x+5=0

(x-5)(x-1)

x=5 or x=1

Check:

√4(5)+5 - √2(5)-1= 2

√25 - √9= 2

5 -3= 2

Therefore 2 work out and is the correct solution and 1 is the extraneous solution.

5r/ r²-9= 2

5r= 2x²-18

(multiply both sides by r-9 to get ride of it on the left)

0= 2x²-5x-18

(2x-9)(x+2)

x= 9/2 or x=-2

Check:

5(-2)/(-2)-9= 2

-10/-5= 2

2= 2

Therefore 2 is the correct answer and 9/2 is the extraneous solution.

x cannot equal 0, -4 because you cannot divide by 0!

1/x + 3/x+4= 2(x(x+4))

1(x+4)/x+4(x) + 3(x)/x+4 (x)= 2(x(x+4))

x+4 +3x= 2x²+8x

4x+4= 2x²+8x

0= 2x²-4x-4

0= x²+2x-2

x= -b ± √b² - 4ac/ 2(b)

x= -2 ± √4-4(1)(-2)/2(2)

x= -2 ± √12/2

x= -2 ± 2√3/2

x= -1 ± √3

*REMINDER: state restrictions before doing anything when dealing with fractions with vaiables in the numerator.

We'll go into more depth tomorrow about absolute numbers, but what we breifly discussed today is:

|5|= 5

|-5|= 5

|x|= 5

x= ±5

|x+1|= 2

case 1: x+1= -2

x= -3

case 2: x+1= 2

x= 1

Check:

|-3 +1| =2

|-2|= 2

|1 + 1|= 2

|2|= 2

Both solutions are possible.

|3|= -3

This is not applicable (N.A) because any absolute number must produce a solution that is positive.

|x-4|= 2x+1

case 1: x-4= 2x+1

x=2x+5

x=-5

case 2: x-4= -(2x+1)

x-4= -2x-1

3x=3

x= 1

Check:

|(-5)-4|= 2(-5)+1

|-9|= -9

5 does not work out, therefore it is an extraneous solution.

Check:

|1+4|= 2(1)+1

|3|= 3

Well that's about all we did in class. I hope this helps. Today we were not assigned any excercise, however, our homework for tonight is already posted up on the blog. Last, but not least, the next scribe......hmmmm who will it be next? I say, the SHEEP of the class is, THAT MEANS YOU NATNAEL!!!! muhahaha (evil laugh) take that sheep. I feel a hard class coming up tomorrow!

-SAMUS

## October 26, 2006

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ReplyDeleteDont call people names like SHEEP =P

ReplyDeleteSam?!?!?!?!?!?!

ReplyDeleteI can't belive this!

what are you guys talking about?

ReplyDeleteben...its something that i, my self couldnt believe:(

ReplyDeleteNat