course Mth 158 ºÏPÝ—é–ÆŸ’ùók‡í®òñd¥²Þassignment #011
......!!!!!!!!...................................
13:15:23 1.2.5. Explain, step by step, how you solved the equation z^2 - z - 6 = 0 using factoring.
......!!!!!!!!...................................
RESPONSE --> first simplify z(-z-6)=0 (z+2)(z-3)=0 (-2,3) confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
13:15:41 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok ------------------------------------------------ Self-critique Rating:ent: 0
.................................................
......!!!!!!!!...................................
13:21:49 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}
......!!!!!!!!...................................
RESPONSE --> factors to (v+1)(v+6)=0 for v+1=0 v must be negative -1 for v+6=0, must be -6 v=(-1,-6) ------------------------------------------------ Self-critique Rating:ent: 2
.................................................
......!!!!!!!!...................................
13:25:09 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}
......!!!!!!!!...................................
RESPONSE --> not sure why answer came up before I put mine in? ------------------------------------------------ Self-critique Rating:ent: 0
.................................................
......!!!!!!!!...................................
13:48:14 1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring.
......!!!!!!!!...................................
RESPONSE --> X^2 + 4x=12 x^2 + 4x +12=12-12 x^2 + 4x +12=0 (x+2x)^2=0 confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
13:50:08 ** 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} **
......!!!!!!!!...................................
RESPONSE --> I was trying a different way and got confused, makes sense now. ------------------------------------------------ Self-critique Rating:ent: 0
.................................................
......!!!!!!!!...................................
13:54:32 1.2.26 (was 1.2.38) (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring.
......!!!!!!!!...................................
RESPONSE --> Multiply x to both sides x + 12/x * x =7x x^2 +12 = 7x subtract 7x from both sides x^2 - 7x + 12= 7x-7x x^2 -7x + 12=0 (X-3)=0 (x-4)=0 x=3, x=4 (3,4) confidence rating: 4 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
13:59:07 ** (x + 2)^2 = 1 so that x + 2 = ± sqrt(1) giving us x + 2 = 1 or x + 2 = -1 so that x = {-1, -3} **
......!!!!!!!!...................................
RESPONSE --> Not sure why showed answer and not problem... here was my work x+2 = sq rt of 1 x+2=-1 and x+2=1 so x could be a -1 or a -3 ------------------------------------------------ Self-critique Rating:ent: 3
.................................................
......!!!!!!!!...................................
14:06:56 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.
......!!!!!!!!...................................
RESPONSE --> x^2 + 2/3x=1/3 x^2 + 2/3x * 1/3=1/3 * 1/3 x^2 + 2/3x * 1/3= 1/9 this is as far as i got confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
14:08:58 ** 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**
......!!!!!!!!...................................
RESPONSE --> that makes sense after reading your notes. ------------------------------------------------ Self-critique Rating:ent: 0
.................................................
......!!!!!!!!...................................
14:14:16 1.2.50 (was 1.2.52) (was 1.3.42). Explain how you solved the equation x^2 + 6x + 1 = 0 using the quadratic formula.
......!!!!!!!!...................................
RESPONSE --> first sub a,b,c and write in formula ax^2 + bx + c=0 a=1 b=6 and c=1 next use b^2-4ac = 6^2 -4(1)(1)=36-4=32 6 +/- sq rt 32/2 confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
14:17:25 ** Starting with x^2 + 6x + 1 = 0 we identify this as having the form a x^2 + b x + c = 0 with a = 1, b = 6 and c = 1. We plug values into quadratic formula to get x = [-6 ± sqrt(6^2 - 4 * 1 * 1) ] / 2 *1 x = [ -6 ± sqrt(36 - 4) / 2 x = { -6 ± sqrt (32) ] / 2 36 - 4 = 32, so x has 2 real solutions, x = { [-6 + sqrt(32) ] / 2 , [-6 - sqrt(32) ] / 2 } Note that sqrt(32) simplifies to sqrt(16) * sqrt(2) = 4 sqrt(2) so this solution set can be written { [-6 + 4 sqrt(2) ] / 2 , [-6 - 4 sqrt(2) ] / 2 }, and this can be simplified by dividing numerators by 2: { -3 + 2 sqrt(2), -3 - 2 sqrt(2) }. **
......!!!!!!!!...................................
RESPONSE --> ok ------------------------------------------------ Self-critique Rating:ent: 0
.................................................
......!!!!!!!!...................................
14:17:33 1.2.74 (was 1.2.72) (was 1.3.66). Explain how you solved the equation pi x^2 + 15 sqrt(2) x + 20 = 0 using the quadratic formula and your calculator.
......!!!!!!!!...................................
RESPONSE --> confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
14:17:38 ** Applying the quadratic formula with a = pi, b = 15 sqrt(2) and c = 20 we get x = [ (-15sqrt(2)) ± sqrt ((-15sqrt(2))^2 -4(pi)(20)) ] / ( 2 pi ). (-15sqrt(2))^2 - 4(pi)(20) = 225 * 2 - 80 pi = 450 - 80 pi = 198.68 approx., so there are 2 unequal real solutions. Our expression is therefore x = [ (-15sqrt(2)) ± 198.68] / ( 2 pi ). Evaluating with a calculator we get x = { 5.62, 1.13 }. DER**
......!!!!!!!!...................................
RESPONSE --> ok ------------------------------------------------ Self-critique Rating:ent:
.................................................
"