12

course MTH 158

I am taking the first two tests Wednesday.

Ÿy¹æÄâÒž‘øê¦xýž·¨w|}a¸Ðòé‡÷µassignment #012

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012. `query 12

College Algebra

02-17-2007

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17:15:43

1.2.5. Explain, step by step, how you solved the equation z^2 - z - 6 = 0 using factoring.

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RESPONSE -->

z^2 - z - 6 = 0

= z^2 - z = 6

This is as far as i could get using my math skills.

using basic trial and error i know

z=3 because

3^2 = 9 -3 = 6 -6=0

confidence assessment: 2

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assignment #012

012. `query 12

College Algebra

02-17-2007

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17:28:11

1.2.5. Explain, step by step, how you solved the equation z^2 - z - 6 = 0 using factoring.

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RESPONSE -->

z^2 - z - 6 = 0

this factors to (z +2)(z-3)

= z +2 =0 and z-3 =0 so

z = -2 and z= 3

confidence assessment: 2

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17:32:55

** STUDENT SOLUTION WITH INSTRUCTOR COMMENT:

I factored this and came up with

(z + 2)(z - 3) = 0

Which broke down to

z + 2 = 0 and z - 3 = 0

This gave me the set {-2, 3}

-2 however, doesn't check out, but only 3 does, so the solution is:

z = 3

INSTRUCTOR COMMENT: It's good that you're checking out the solutions, but note that -2 also checks: (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0. **

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RESPONSE -->

I dont know if it maters but i restarted this section because i felt i was in the wrong spot.

I very much understand my mistakes with this problem.

self critique assessment: 2

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18:17:05

1.1.72. Explain, step by step, how you solved the equation x^3 + 6 x^2 - 7 x = 0 using factoring.

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RESPONSE -->

x^3 + 6 x^2 - 7 x = 0

=x( x^2 + 6x - 7)+ 0 then

x( x - 1 ) ( x + 7 )

confidence assessment: 2

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18:21:00

** Starting with

x^3 + 6 x^2 - 7 x = 0 factor x out of the left-hand side:

x(x^2 + 6x - 7) = 0. Factor the trinomial:

x ( x+7) ( x - 1) = 0. Then

x = 0 or x + 7 = 0 or x - 1 = 0 so

x = 0 or x = -7 or x = 1. **

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RESPONSE -->

Can x ( x+7) ( x - 1) = 0. also be written

x ( x - 1) ( x+7) = 0.

self critique assessment: 2

The order doesn't matter--your factoring was fine.

Be sure you understand how this gives you the solutions of the equation, which you didn't include.

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18:29:40

1.2.14 (was 1.3.6). Explain how you solved the equation by factoring.

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RESPONSE -->

Thair is no problem listed so i will do # 14

I just reolized that i dont know how to wright this. i will try

first bring the -1^5 to get

2x - 1 = 1 then

2x= 2

=x = 1

confidence assessment: 2

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18:31:03

STUDENT SOLUTION: v^2+7v+6=0. This factors into

(v + 1) (v + 6) = 0, which has solutions

v + 1 = 0 and v + 6 = 0, giving us

v = {-1, -6}

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RESPONSE -->

I dont know whar this problem came from.

self critique assessment: 2

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18:39:11

1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring.

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RESPONSE -->

x(x+4)=12 first lose ( )

x^2 + 4x = 12 then -12

x^2 + 4x - 12 = 0 then factor

= ( x + 6 ) ( x - 2 )

confidence assessment: 2

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18:40:57

** Starting with

x(x+4)=12 apply the Distributive Law to the left-hand side:

x^2 + 4x = 12 add -12 to both sides:

x^2 + 4x -12 = 0 factor:

(x - 2)(x + 6) = 0 apply the zero property:

(x - 2) = 0 or (x + 6) = 0 so that

x = {2 , -6} **

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RESPONSE -->

i Knew that i just didnt go far enough

self critique assessment: 2

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19:21:49

1.2.38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring.

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RESPONSE -->

x + 12/x = 7 I cant seem to figure this out .

the 7 needs to come over so i get x + 12/x - 7 =0

but what to do with the 12/x the only thing could be

to multiply the x = x^2 + 12 - 7 =0 but this isnt complete x^2 + 12 ? - 7 =0

confidence assessment: 0

Right idea, but it will be -7x, not just -7

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19:23:21

** Starting with

x + 12/x = 7 multiply both sides by the denominator x:

x^2 + 12 = 7 x add -7x to both sides:

x^2 -7x + 12 = 0 factor:

(x - 3)(x - 4) = 0 apply the zero property

x-3 = 0 or x-4 = 0 so that

x = {3 , 4} **

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RESPONSE -->

now i feel dumb

self critique assessment: 2

Not in the least. You had the right idea, just missed a single detail.

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19:38:55

1.2.32 (was 1.3.24). Explain how you solved the equation (x+2)^2 = 1 by the square root method.

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RESPONSE -->

(x+2)^2 = 1 using (x + a)^2 = x^2 + 2ax + a^2

I get x^2 + 4x + 4 = 1 then subtract 1

= x^2 + 4x + 3 = 0 now factor

( x + 1 ) ( x + 3 ) then

= x + 1 = 0 and x + 3 = 0 so

x = -1 and x = -3

confidence assessment: 2

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19:40:51

** (x + 2)^2 = 1 so that

x + 2 = ± sqrt(1) giving us

x + 2 = 1 or x + 2 = -1 so that

x = {-1, -3} **

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RESPONSE -->

I still got the answer.

self critique assessment: 2

Good solution, but be sure you understand the square root method.

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20:07:52

1.2.44 (was 1.3.36). Explain how you solved the equation x^2 + 2/3 x - 1/3 = 0 by completing the square.

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RESPONSE -->

x^2 + 2/3 x - 1/3 = 0

=x^2 + 2/3 x = 1/3

=x^2 + 2/3 x + 1/9 = 1/3 + 1/9

=(x + 1/3)^2 = 4/9

= x + 1/3= 2/3

= x + 1/3= 2/3

= x = 1/3

confidence assessment: 2

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20:09:12

** x^2 + 2/3x - 1/3 = 0. Multiply both sides by the common denominator 3 to get

3 x^2 + 2 x - 1 = 0. Factor to get

(3x - 1) ( x + 1) = 0. Apply the zero property to get

3x - 1 = 0 or x + 1 = 0 so that

x = 1/3 or x = -1.

DER**

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RESPONSE -->

i am geting it.

self critique assessment: 2

I believe you are.

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20:25:52

1.2.52 (was 1.3.42). Explain how you solved the equation x^2 + 6x + 1 = 0 using the quadratic formula.

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RESPONSE -->

confidence assessment: 0

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"

You did well here. Missed a few details, but you got most of them and you understand the process. Let me know if you have questions.