012. `* 12 * 1.4.12 (was 1.4.6). Explain how you found the real solutions of the equation (1-2x)^(1/3) - 1 = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since -1 does not contain the variable to solve for, move it to the right hand side of the equation by adding 1 to both sides. Get rid of the fraction exponent on the lefthand side by raising both sides to the 3 power...((1-2x)^1/2)^3=(1)^3 Factor...(1-2x))=(1)^3 Remove the parentheses around the expression 1-2x...1-2x=(1)^3...1-2x=1. Since 1 does not have the variable to solve for, subtract 1 from boh sides..-2x=-1+ 1..-2x=0...divide x=0/-2..x=0 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with (1-2x)^(1/3)-1=0 add 1 to both sides to get (1-2x)^(1/3)=1 then raise both sides to the power 3 to get [(1-2x)^(1/3)]^3 = 1^3. Since [(1-2x)^(1/3)]^3 = (1 - 2x) ^( 1/3 * 3) = (1-2x)^1 = 1 - 2x we have 1-2x=1. Adding -1 to both sides we get -2x=0 so that x=0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 3 OK ********************************************* Question: * 1.4.28 (was 1.4.18). Explain how you found the real solutions of the equation sqrt(3x+7) + sqrt(x+2) = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sqrt(3x+7)= -sqrt(x+2) + 1 . Now we square both sides to get sqrt(3x+7)^2 =[ -sqrt(x+2) +1]^2. Expanding the right-hand side using (a+b)^2 = a^2 + 2 a b + b^2 with a = -sqrt(x+2) and b = 1: 3x+7= x+2 - 2sqrt(x+2) +1. Note that whatever we do we can't avoid that term -2 sqrt(x+2). Simplifying 3x+7= x+ 3 - 2sqrt(x+2) then adding -(x+3) we have 3x+7-x-3 = -2sqrt(x+2). Squaring both sides we get (2x+4)^2 = (-2sqrt(x+2))^2. Note that when you do this step you square away the - sign. This can result in extraneous solutions. We get 4x^2+16x+16= 4(x+2). Applying the distributive law we have 4x^2+16x+16=4x+8. Adding -4x - 8 to both sides we obtain 4x^2+12x+8=0. Factoring 4 we get 4*((x+1)(x+2)=0 and dividing both sides by 4 we have (x+1)(x+2)=0 Applying the zero principle we end up with (x+1)(x+2)=0 so that our potential solution set is x= {-1, -2}. Both of these solutions need to be checked in the original equation sqrt(3x+7)+sqrt(x+2)=1 the solution -2 gives us sqrr(1) + sqrt(0) = 1 or 1 + 0 = 1, which is true. x = -1 is an extraneous solution that was introduced in our squaring step. Thus our only solution is x = -2. confidence rating #$&*: 3 OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. COMMON ERROR sqrt(3x+7)^2= (-sqrt(x+2)+1)^2 (3x+7)=(x+2)+1 EXPLANATION (a + b)^2 = (a + b) * (a + b) = a ( a + b) + b ( a + b) = a * a + a * b + b * a + b * b. So (a + b)^2 is not equal to a^2 + b^2. Nor is it so that (-sqrt(x+2)+1)^2 is equal to (x+2) + 1. When you square the right-hand side you can't write it -sqrt(x+2)+1 ^2 because written this way only the 1 gets squared. You have to write it (-sqrt(x+2)+1) ^2, which means (-sqrt(x+2) + 1) * (-sqrt(x+2) + 1). When multiplied out using the distributive law, using the same sequence of steps used above to find (a + b) ( a + b), you should get (x + 2) - 2 sqrt( x + 2) + 1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 3 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: Find a common factor for each term. 1(x^1/4)^3 minus 9x^1/4=0...Substitute u=x^1/4 and solve for u. u^3-9u=0 where u=x^1/4. Factor out the GCF of u from each term in the polynomial u(u^2)+u(-9)=0. Factor out the GCF of u from u^3-9u...u(u^2-9)=0...The binomial can be factored using the difference of squares formula, because both terms are perfect squares..Where a=u and b=3 u(u-3)(u+3)=0 Set the single term factor on the left hand ide of the equation = to zero … u=0,3,-3 Substitute the x^1/4=u back into the solved equation in order to solve for x...x^1/4=0..Solve equation for x then we put x^1/4=3 then so x=81. X^1/4=-3..x=81...0,81,81 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Here we can factor x^(1/4) from both sides: Starting with x^(3/4) - 9 x^(1/4) = 0 we factor as indicated to get x^(1/4) ( x^(1/2) - 9) = 0. Applying the zero principle we get x^(1/4) = 0 or x^(1/2) - 9 = 0 which gives us x = 0 or x^(1/2) = 9. Squaring both sides of x^(1/2) = 9 we get x = 81. So our solution set is {0, 81). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 3 OK ********************************************* Question: * 1.4.46 (was 1.4.36). Explain how you found the real solutions of the equation x^6 - 7 x^3 - 8 =0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a = x^3. Then a^2 = x^6 and the equation x^6 - 7x^3 - 8=0 becomes a^2 - 7 a - 8 = 0. This factors into (a-8)(a+1) = 0, with solutions a = 8, a = -1. Since a = x^3 the solutions are x^3 = 8 and x^3 = -1. We solve these equations to get x = 8^(1/3) = 2 and x = (-1)^(1/3) = -1. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: 3 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: Alright u = sqrt(x^2 - 3x). Then u^2 = x^2 - 3x, and the equation is u^2 - u = 2. u^2 - u - 2 = 0. (u-2)(u+1) = 0. Solutions are u = 2, u = -1. sqrt(x^2 - 3 x) = 2 sqrt(x^2 - 3 x) = -1. sqrt(x^2 - 3x) = 2 x^2 - 3x = 4. x^2 - 3x - 4 = 0 (x-4)(x+1) = 0 x = 4 or x = -1. confidence rating #$&*: 3 OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): 3 OK ------------------------------------------------ Self-critique Rating: ********************************************* 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: Well x^4 plus radical2x^2 minus 2 =0. We move the term after the radical to the front of the radical (radical 2x^2) x^4 plus times x^2 radical 2 minus 2= 0. Multiply x^2 to get x^2 x^4 plus times x^2 radical 2 minus 2= 0 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: 3 OK Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!