#$&* course Mth 158 11:00AM 2/22/13 012. `* 12
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Given Solution: * * Starting with (1-2x)^(1/3)-1=0 add 1 to both sides to get (1-2x)^(1/3)=1 then raise both sides to the power 3 to get [(1-2x)^(1/3)]^3 = 1^3. Since [(1-2x)^(1/3)]^3 = (1 - 2x) ^( 1/3 * 3) = (1-2x)^1 = 1 - 2x we have 1-2x=1. Adding -1 to both sides we get -2x=0 so that x=0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 1.4.28 (was 1.4.18). Explain how you found the real solutions of the equation sqrt(3x+7) + sqrt(x+2) = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sqrt(3x+7) + sqrt(x+2) = 1 Isolate sqrt(3x+7): sqrt(3x+7) = 1 - sqrt(x+2) Square both sides: (sqrt(3x+7))^2 = (1 - sqrt(x+2))^2 3x + 7 = (1 - sqrt(x+2)(1 - sqrt(x+2) 3x + 7 = 1 - sqrt(x+2) - sqrt(x+2) + (sqrt(x+2))^2 3x + 7 = 1 - 2sqrt(x+2) + x + 2 3x + 7 = 3 + x - 2sqrt(x+2) Add -3 and -x to both sides: 2x + 4 = -2sqrt(x+2) Square both sides: (2x + 4)^2 = (-2sqrt(x+2))^2 4x^2 + 16x + 16 = 4x + 8 Add -4x and - 8 to both sides: 4x^2 + 12x + 8 = 0 Factor out 4: 4(x^2 + 3x + 2) = 0 4(x + 1)(x + 2) = 0 Zero product property: x = -1 or x = -2 Check: x = -1 sqrt(3(-1)+7) + sqrt(-1+2) = 1 sqrt(4) + sqrt(1) = 1 2 + 1 = 1 3 ≠ 1 Check x = -2 sqrt(3(-2)+7) + sqrt(-2+2) = 1 sqrt(1) + sqrt(0) = 1 1 = 1 Solution Set: {-2} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with sqrt(3x+7)+sqrt(x+2)=1 we could just square both sides, recalling that (a+b)^2 = a^2 + 2 a b + b^2. This would be valid but instead we will add -sqrt(x+2) to both sides to get a form with a square root on both sides. This choice is arbitrary; it could be done either way. We get sqrt(3x+7)= -sqrt(x+2) + 1 . Now we square both sides to get sqrt(3x+7)^2 =[ -sqrt(x+2) +1]^2. Expanding the right-hand side using (a+b)^2 = a^2 + 2 a b + b^2 with a = -sqrt(x+2) and b = 1: 3x+7= x+2 - 2sqrt(x+2) +1. Note that whatever we do we can't avoid that term -2 sqrt(x+2). Simplifying 3x+7= x+ 3 - 2sqrt(x+2) then adding -(x+3) we have 3x+7-x-3 = -2sqrt(x+2). Squaring both sides we get (2x+4)^2 = (-2sqrt(x+2))^2. Note that when you do this step you square away the - sign. This can result in extraneous solutions. We get 4x^2+16x+16= 4(x+2). Applying the distributive law we have 4x^2+16x+16=4x+8. Adding -4x - 8 to both sides we obtain 4x^2+12x+8=0. Factoring 4 we get 4*((x+1)(x+2)=0 and dividing both sides by 4 we have (x+1)(x+2)=0 Applying the zero principle we end up with (x+1)(x+2)=0 so that our potential solution set is x= {-1, -2}. Both of these solutions need to be checked in the original equation sqrt(3x+7)+sqrt(x+2)=1 As it turns out: the solution -1 gives us sqrt(4) + sqrt(1) = 1 or 2 + 1 = 1, which isn't true, while the solution -2 gives us sqrr(1) + sqrt(0) = 1 or 1 + 0 = 1, which is true. x = -1 is an extraneous solution that was introduced in our squaring step. Thus our only solution is x = -2. ** STUDENT QUESTION I got to the third step but I got confused on what to eliminate or substitute in, looking at the solution, im still a little confused on how it all worked out. U got any suggestions on how to look at it in a better way??? INSTRUCTOR RESPONSE You're pretty much stuck with this technique and this way of looking at the problem. It should be pretty clear to you that (sqrt(x+3))^2 is just x + 3. Squaring the expression [ -sqrt(x+2) +1] is a little more challenging. We could use the distributive law: [ -sqrt(x+2) +1]^2 = [-sqrt(x + 2) + 1 ] * [-sqrt(x + 2) + 1 ] = -sqrt(x+2) * [-sqrt(x + 2) + 1 ] + 1 * [-sqrt(x + 2) + 1 ] = -sqrt(x+2) * (-sqrt(x + 2) ) + (-sqrt(x + 2) * 1 + 1 * (-sqrt(x + 2) ) + 1 * 1= (x + 2) - sqrt(x + 2) - sqrt(x + 2) + 1= x+2 - 2sqrt(x+2) +1. Once we get the equation 3x+7= x+2 - 2sqrt(x+2) +1 we see that we still need to 'get to' that x within the square root. To do that we rearrange the equation so that the square root is on one side, all by itself, so we can square it without dragging a lot of other stuff along. So we do a couple of steps and we get 3x+7-x-3 = -2sqrt(x+2). If we square both sides of this equation, we get rid of all the square roots and we get x out where we can deal with it. The details are in the given solution, but we get the equation 4x^2+16x+16= 4(x+2). This equation now has x^2 and x terms, so we know it's a quadratic, and we rearrange and solve it as such. The details are in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 1.4.40 (was 1.4.30). Explain how you found the real solutions of the equation x^(3/4) - 9 x^(1/4) = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^(3/4) - 9 x^(1/4) = 0 Add 9 x^(1/4) to both sides: x^(3/4) = 9 x^(1/4) Raise each side by the power of 4: (x^(3/4))^4 = (9 x^(1/4))^4 x^3 = 9x
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Given Solution: * * Here we can factor x^(1/4) from both sides: Starting with x^(3/4) - 9 x^(1/4) = 0 we factor as indicated to get x^(1/4) ( x^(1/2) - 9) = 0. Applying the zero principle we get x^(1/4) = 0 or x^(1/2) - 9 = 0 which gives us x = 0 or x^(1/2) = 9. Squaring both sides of x^(1/2) = 9 we get x = 81. So our solution set is {0, 81). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay, looking at the given solution, I can see it got its answer and was able to work it myself. Where did I go wrong with my work???? ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 1.4.46 (was 1.4.36). Explain how you found the real solutions of the equation x^6 - 7 x^3 - 8 =0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^6 - 7 x^3 - 8 = 0 u = x^3 u^2 - 7u - 8 = 0 (u - 8)(u + 1) = 0 Zero product property: u = 8, or, u = -1 u = x^3 = 8 = 2, or, u = x^3 = -1 Check: x = 2 (2)^6 - 7 (2)^3 - 8 = 0 64 - 56 - 8 = 0 0 = 0 Check: x = -1 (-1)^6 - 7 (-1)^3 - 8 = 0 1 + 7 - 8 = 0 0 = 0 Solution Set: {2 , -1} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Let a = x^3. Then a^2 = x^6 and the equation x^6 - 7x^3 - 8=0 becomes a^2 - 7 a - 8 = 0. This factors into (a-8)(a+1) = 0, with solutions a = 8, a = -1. Since a = x^3 the solutions are x^3 = 8 and x^3 = -1. We solve these equations to get x = 8^(1/3) = 2 and x = (-1)^(1/3) = -1. STUDENT QUESTION I am confused as to why you substituted the a. I know how to do this on the calculator by using y = and 2nd graph to get the solution (-1, 2) INSTRUCTOR RESPONSE If you substitute a for x^3, then you end up with a quadratic equation that can be easily factored. If a = x^3, then x^6 = a^2 so the equation becomes a^2 - 7 a - 8 = 0. We factor this and find that a can be either 8 or -1. So x^3 can be either 8 or -1. Thus x can be either 2 or -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 1.4.64 (was 1.4.54). Explain how you found the real solutions of the equation x^2 - 3 x - sqrt(x^2 - 3x) = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 - 3 x - sqrt(x^2 - 3x) = 2 Isolate - sqrt(x^2 - 3x): - sqrt(x^2 - 3x) = - x^2 - 3 x + 2 Square both sides: (- sqrt(x^2 - 3x))^2 = (- x^2 - 3 x + 2)^2 x^2 - 3x = x^4 -3x^3 - 2x - 3x^3 + 9x^2 + 6x - 2x^2 + 6x + 4 x^2 - 3x = x^4 - 6x^3 + 7x^2 + 10x + 4 - x^4 + 6x^3 - 6x^2 - 13x - 4 = 0 I'm not sure where to go from here. I tried factoring by grouping: (- x^4 + 6x^3) + (- 6x^2 - 13x - 4) = 0 -x^3(x - 6) + (- 6x^2 - 13x - 4) = 0 But I don't know where to take (- 6x^2 - 13x - 4). confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Let u = sqrt(x^2 - 3x). Then u^2 = x^2 - 3x, and the equation is u^2 - u = 2. Rearrange to get u^2 - u - 2 = 0. Factor to get (u-2)(u+1) = 0. Solutions are u = 2, u = -1. Substituting x^2 - 3x back in for u we get sqrt(x^2 - 3 x) = 2 and sqrt(x^2 - 3 x) = -1. The second is impossible since sqrt can't be negative. The first gives us sqrt(x^2 - 3x) = 2 so x^2 - 3x = 4. Rearranging we have x^2 - 3x - 4 = 0 so that (x-4)(x+1) = 0 and x = 4 or x = -1. STUDENT QUESTION I got stuck on this part u=(-sqrt2+-sqrt10)/2, but after I looked at the solution, it made a little more sense to me, but im not real confident. Got any suggestions on how to approach it in a different way??? INSTRUCTOR RESPONSE Plugging into the quadratic formula we get u=(-sqrt2+-sqrt10)/2, meaning u can take one of the two values u=(-sqrt2+sqrt10)/2 or u=(-sqrt2-sqrt10)/2. These quantities are just plain old numbers, which you could evaluate (up to some roundoff) on your calculator. The first possible value of u is about equal to about .874. The second possible value of u is negative. Now u stands for x^2, so we ignore the negative value of u (this since x^2 can't be negative). So we're left with x^2 = u = .874. So x = +- sqrt(.874), giving us the values of x in the given solution. STUDENT QUESTION I still do not understand using u. I can do it from the 2nd step. Problem: u^2 - u - 2 = 0. Factor: (u-2)(u + 1) = 0. You get u = 2, -1. You will solve x^2 - 3x -4. Factor will be (x - 4)(x + 1) = 0. Solutions are 4, -1. INSTRUCTOR RESPONSE The left-hand side consists of x^2 - 3x and the square root of x^2 - 3x. So instead of x^2 - 3 x - sqrt(x^2 - 3x) we write the left-hand side as u - sqrt(u), which is easier to deal with. We solve for u, then come back and figure out what value(s) of x give us our values of u. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay, I see how to get the answer and I worked it through it myself. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 1.4.92 \ 90 (was 1.4.66). Explain how you found the real solutions of the equation x^4 + sqrt(2) x^2 - 2 = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I'm not sure how to work this problem. I tried isolating the square root: x^4 + sqrt(2) x^2 - 2 = 0 (sqrt(2))^2 = [(-x^4 + 2) / x^2]^2 2 = (x^8 + 4) / x^4 And I tried solving in quadratic form: x^4 + sqrt(2) x^2 - 2 = 0 u = x^2 u^2 + sqrt(2)u - 2 = 0 But am not sure how to factor that. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^4+ sqrt(2)x^2-2=0 we let u=x^2 so that u^2 = x^4 giving us the equation u^2 + sqrt(2)u-2=0 Using the quadratic formula u=(-sqrt2 +- sqrt(2-(-8))/2 so u=(-sqrt2+-sqrt10)/2 Note that u = (-sqrt(2) - sqrt(10) ) / 2 is negative, and u = ( -sqrt(2) + sqrt(10) ) / 2 is positive. u = x^2, so u can only be positive. Thus the only solutions are the solutions to the equation come from x^2 = ( -sqrt(2) + sqrt(10) ) / 2. The solutions are x = sqrt( ( -sqrt(2) + sqrt(10) ) / 2 ) and x = -sqrt( ( -sqrt(2) + sqrt(10) ) / 2 ). Approximations to three significant figures are x = .935 and x = -.935. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don't understand how we go from: u^2 + sqrt(2)u-2=0 to u=(-sqrt2 +- sqrt(2-(-8))/2. Using the quadratic formula. Could you explain this please???? I understand how to get the solutions after u=(-sqrt2 +- sqrt(2-(-8))/2, but don't understand how to get there. ------------------------------------------------ Self-critique Rating: 1