#$&* course mth 158 hard lesson If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENT: I factored this and came up with (z + 2)(z - 3) = 0 Which broke down to z + 2 = 0 and z - 3 = 0 This gave me the set {-2, 3} -2 however, doesn't check out, but only 3 does, so the solution is: z = 3 INSTRUCTOR COMMENT: It's good that you're checking out the solutions, because sometimes we get extraneous roots. But note that -2 also checks: (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.14 (was 1.3.6). Explain how you solved the equation v^2+7v+6=0 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v^2+7v+6=0 (v + 1) (v + 6) = 0 v + 1 = 0 and v + 6 = 0 v = {-1, -6} confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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} &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x(x+4)=12 apply the Distributive Law to the left-hand side: x^2 + 4x = 12 add -12 x^2 + 4x -12 = 0 (x - 2)(x + 6) = 0 (x - 2) = 0 or (x + 6) = 0 x = {2 , -6} confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x(x+4)=12 apply the Distributive Law to the left-hand side: x^2 + 4x = 12 add -12 to both sides: x^2 + 4x -12 = 0 factor: (x - 2)(x + 6) = 0 apply the zero property: (x - 2) = 0 or (x + 6) = 0 so that x = {2 , -6} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.26 \ 38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x + 12/x = 7 x^2 + 12 = 7 x x^2 -7x + 12 = 0 (x - 3)(x - 4) = 0 x-3 = 0 or x-4 = 0 x = 3 , 4 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.32 (was 1.3.24). Explain how you solved the equation (x+2)^2 = 1 by the square root method. * * (x + 2)^2 = 1 x + 2 = ± sqrt(1) x + 2 = 1 or x + 2 = -1 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1. Given Solution: * * (x + 2)^2 = 1 so that x + 2 = ± sqrt(1) giving us x + 2 = 1 or x + 2 = -1 so that x = {-1, -3} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Missed the last step ok ------------------------------------------------ Self-critique Rating:
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Given Solution: * * 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. STUDENT QUESTION: The only thing that confuses me is the 1/3. Is that because of the 3x? INSTRUCTOR RESPONSE: You got the equation (3x - 1) ( x + 1) = 0. The product of two numbers can be zero only if one of the numbers is zero. So (3x - 1) ( x + 1) = 0 means that 3x - 1 = 0 or x + 1 = 0. You left out this step in your solution. x + 1 = 0 is an equation with solution x = -1 Thus the solution to our original equation is x = 1/3 or x = -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok fuzzy ------------------------------------------------ Self-critique Rating:
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Given Solution: * * To find the clock time at which the object is 15 m off the ground we set the height s equal to 15 to get the equation -4.9t^2 + 20t = 15 Subtracting 15 from both sides we get -4.9t^2 +20t - 15 = 0 so that t = { -20 ± sqrt [20^2 - 4(-4.9)(-15) ] } / (2(-4.9)) Numerically these simplify to t = .99 and t = 3.09. Interpretation: • The object passes the 15 m height on the way up at t = 99 and again on the way down at t = 3.09. To find when the object strikes the ground we set s = 0 to get the equation -4.9t^2 + 20t = 0 which we solve to get t = [ -20 ± sqrt [20^2 - 4(-4.9)(0)] ] / (2(-4.9)) This simplifies to t = [ -20 +-sqrt(20^2) ] / (2 * -4.9) = [-20 +- 20 ] / (-9.8). The solutions simplify to t = 0 and t = 4.1 approx. Interpretation: The object is at the ground at t = 0, when it starts up, and at t = 4.1 seconds, when it again strikes the ground. To find when the altitude is 100 we set s = 100 to get -4.9t^2 + 20t = 100. Subtracting 100 from both sides we obtain -4.9t^2 +20t - 100 = 0 which we solve using the quadratic formula. We get t = [ -20 ± sqrt (20^2 - 4(-4.9)(-100)) ] / (2(-4.9)) The discriminant is 20^2 - 4 * -4.9 * -100, which turns out to be negative so we do not obtain a solution. Interpretation: We conclude that this object will not rise 100 ft. ** STUDENT QUESTION I was lost on this question and even reading the solution, Im still confused about it. Do you have any suggestions on how to look at it in a different way??? INSTRUCTOR RESPONSE s = -4.9 t^2 + 20 t means that if you plug in a value of t, you get the height, which is represented by the variable s. Also if you want to find the value of t that gives you a certain height, you plug in that height for s and solve the equation for t. The first question asks you to find when the height is 15 meters. So you plug in 15 for s. What equation do you get? You get the equation 15 = -4.9 t^2 + 20 t. Now you solve the equation for t. How do you do that? The equation is quadratic, since it contains both t^2 and t. The standard form for a quadratic equation is a t^2 + b t + c = 0 In this form you can try to factor the left-hand side. If this is possible you can then apply the zero property, as you've done in some of the preceding problems. If you can't figure out how to factor the equation (and in real-world problems you usually can't), you can use the quadratic formula. In this case you rearrange the equation 15 = -4.9 t^2 + 20 t to the form -4.9 t^2 + 20 t - 15 = 0 and pretty quickly realize that you won't be able to factor it. So you use the quadratic formula, as shown in the given solution. STUDENT QUESTION This is the part I am confused about. How did it go from the top equation to the numbers .99 and 3.09? INSTRUCTOR RESPONSE To evaluate { -20 ± sqrt [20^2 - 4(-4.9)(-15) ] } / (2(-4.9)) follow the order of operations: First evaluate the expression in braces: { -20 ± sqrt [20^2 - 4(-4.9)(-15) ] } = -20 ± sqrt [20^2 - 4(-4.9)(-15) ] Start with the expression in brackets, which is 20^2 - 4(-4.9)(-15) Find 20^2, multiply 4(-4.9)(-15), and subtract the second result from the first. Then take the square root of the result, giving you the value of sqrt(20^2 - 4(-4.9)(-15)). Now evaluate -20 ± sqrt [20^2 - 4(-4.9)(-15) ] You get two results, one based on the + of the ± and the other on the -: First add the value of the square root to -20. Then subtract the value of the square root from -20. This completes evaluation of the quantity in braces. Now evaluate the denominator 2 ( -4.9), which you multiply to get -9.8. Divide the first value of the numerator by -9.8 to get .99. Divide the second value of the numerator by -9.8 to get 3.09. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok