#$&* course Mth 158 3/4 1pm 012. `* 12* 1.4.12 (was 1.4.6). Explain how you found the real solutions of the equation (1-2x)^(1/3) - 1 = 0
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Given Solution: * * Starting with (1-2x)^(1/3)-1=0 add 1 to both sides to get (1-2x)^(1/3)=1 then raise both sides to the power 3 to get [(1-2x)^(1/3)]^3 = 1^3. Since [(1-2x)^(1/3)]^3 = (1 - 2x) ^( 1/3 * 3) = (1-2x)^1 = 1 - 2x we have 1-2x=1. Adding -1 to both sides we get -2x=0 so that x=0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.4.28 (was 1.4.18). Explain how you found the real solutions of the equation sqrt(3x+7) + sqrt(x+2) = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Square both sides of the equation, and we’re left with 3x + 7 + x + 2 = 1 Add like terms 4x + 9 = 1 Subtract 9 from both sides 4x = -8 Divide both sides by 4 X = -2 confidence rating #$&*: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, I worked the problem with ease, but when I saw the solution and the alternative way to solve for x, I couldn’t understand why the more complicated method would be necessary when I got the same answer using a much shorter route. ------------------------------------------------ Self-critique Rating:3
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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):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: To make this problem easier to solve, we can use a substitution method A = x^3 Therefore the problem will now read a^2 + 7a - 8 = 0 factor a^2 + 8a - a - 8 = 0 a(a - 8) + 1(a - 8) = 0 (a + 1)(a - 8) = 0 So a= -1, 8 A = x^3 so X^3 = -1 and x^3 = 8 So, to simplify, multiply both sides by the exponent of ^1/3 to get the solutions X = -1, 2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Let a = x^3. Then a^2 = x^6 and the equation x^6 - 7x^3 - 8=0 becomes a^2 - 7 a - 8 = 0. This factors into (a-8)(a+1) = 0, with solutions a = 8, a = -1. Since a = x^3 the solutions are • x^3 = 8 and • x^3 = -1. We solve these equations to get • x = 8^(1/3) = 2 and • x = (-1)^(1/3) = -1. STUDENT QUESTION I am confused as to why you substituted the a. I know how to do this on the calculator by using y = and 2nd graph to get the solution (-1, 2) INSTRUCTOR RESPONSE If you substitute a for x^3, then you end up with a quadratic equation that can be easily factored. If a = x^3, then x^6 = a^2 so the equation becomes a^2 - 7 a - 8 = 0. We factor this and find that a can be either 8 or -1. So x^3 can be either 8 or -1. Thus x can be either 2 or -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok, I had a little help with the substitution, but it all makes sense so I feel good about it ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.4.64 (was 1.4.54). Explain how you found the real solutions of the equation x^2 - 3 x - sqrt(x^2 - 3x) = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can substitute a for sqrt(x^2 - 3x) to make the equation a little simpler to solve So the equation will now read a^2 - a = 2 subtract 2 from both sides a^2 - a - 2 = 0 factor a^2 - 2a + a - 2 = 0 a(a - 2) + 1(a - 2) = 0 (a + 1)(a - 2) = 0 a= -1, or 2 we know a square root can’t be negative so a = -1 is not possible therefore a = 2 a= sqrt(x^2 - 3x) so sqrt(x^2 - 3x) = 2 square both sides x^2 - 3x = 4 subtract 4 from both sides x^2 - 3x - 4 = 0 factor x^2 - 4x + x - 4 = 0 x(x - 4) + 1(x - 4) = 0 (x + 1)(x - 4) = 0 X = -1 or 4 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: Substitute a for x^2 to make solving the problem a little less complicated Now we have a^2 + sqrt(2)a - 2 = 0 use the quadractic equation [-sqrt(2) ± sqrt((2)^2 - 4(-2))]/2 Simplify [-sqrt(2) ± sqrt(10)]/2 The solution [-sqrt(2) - sqrt(10)]/2 is negative, so it cannot be possible So our solution is a = (-sqrt(2) + sqrt(10)) / 2 a = x^2 so x^2 = [-sqrt(2) + sqrt(10)]/2 so x = ± sqrt[(-sqrt(2) + sqrt(10))/ 2] confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok, this one was a little more difficult, but I feel very comfortable with it ------------------------------------------------ Self-critique Rating:3 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: #*&!