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 February 16th, 2015, 10:51 PM #1 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty Quick question. Just a quick question: $\displaystyle (x-3)^3 - (x+1)(x-3)^2 = 0$ can the above be solved as such: $\displaystyle (x-3)^3 - (x+1)(x-3)^2 = 0$ $\displaystyle (x-3)^3 = (x+1)(x-3)^2$ $\displaystyle (x-3)^3/(x-3)^2 = (x+1)$ $\displaystyle (x-3) = (x+1)$ if so, then could the following be solved the same way? $\displaystyle 2(x-1)^5 - x(x-1)^4 = 0$ $\displaystyle 2(x-1)^5 = x(x-1)^4$ $\displaystyle 2(x-1)^5/(x-1)^4 = x$ $\displaystyle 2(x-1) = x$ Thank you in advance for the feedback.
 February 16th, 2015, 10:55 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs For the first problem: $\displaystyle (x-3)^3-(x+1)(x-3)^2= 0$ I would factor: $\displaystyle (x-3)^2\left((x-3)-(x+1)\right)= 0$ Collect like terms: $\displaystyle 4(x-3)^2= 0$ Hence: $\displaystyle x=3$ Can you apply the same method to the second problem? Thanks from hyperbola
 February 16th, 2015, 11:30 PM #3 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty $\displaystyle 2(x-1)^5 - x(x-1)^4$ common factor is $\displaystyle (x-1)^4$ $\displaystyle (x-1)^4(2(x-1)-x) = 0$ $\displaystyle (x-1)^4(2x-2-x) = 0$ $\displaystyle (x-1)^4(x-2) = 0$ $\displaystyle (x-1)^4 = 0, x=1$ $\displaystyle (x-2) = 0, x=2$ correct? Thanks from MarkFL Last edited by hyperbola; February 16th, 2015 at 11:32 PM.
 February 17th, 2015, 12:48 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs Yes...perfect! Thanks from hyperbola
 February 17th, 2015, 03:44 AM #5 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty MarkFL could you help me out with one more question please? I'm asked to solve the following by method of 'completing the square'. $\displaystyle 2x^2 + 12x + 3 = 0$ $\displaystyle 2x^2+12x=-3$ $\displaystyle (0.5 x 12)^2=36$ $\displaystyle 2x^2+12x+36=-3+36$ $\displaystyle 2x^2+12x+36=33$ $\displaystyle 2[x^2+6x+18]=33$ No common factors here. As far as I can see, this can only be solved using the quadratic formula. Using quadratic formula I get $\displaystyle x1 = -0.26$ $\displaystyle x2 = -5.74$ Last edited by hyperbola; February 17th, 2015 at 04:38 AM.
 February 17th, 2015, 09:14 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs For future reference, it is really preferable that you begin a new thread for a new question...this way threads do not become possibly convoluted and hard to follow. We are given: $\displaystyle 2x^2+12x+3=0$ Subtract through by 3: $\displaystyle 2x^2+12x=-3$ Divide through by 2: $\displaystyle x^2+6x=-\frac{3}{2}$ Take the coefficient of the linear term, which is 6, divide it by 2 to get 3, then square that to get 9, and so add 9 to both sides: $\displaystyle x^2+6x+9=\frac{15}{2}$ Rewrite the LHS as a square: $\displaystyle (x+3)^2=\frac{15}{2}$ Use the square root property: $\displaystyle x+3=\pm\sqrt{\frac{15}{2}}=\pm\frac{\sqrt{30}}{2}$ Subtract through by 3: $\displaystyle x=\frac{-6\pm\sqrt{30}}{2}$ Thanks from hyperbola
February 17th, 2015, 11:29 AM   #7
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Quote:
 Originally Posted by hyperbola $\displaystyle (x-3)^3 = (x+1)(x-3)^2$ $\displaystyle (x-3)^3/(x-3)^2 = (x+1)$
This division is only valid when $x \ne 3$, otherwise you are dividing by zero. In this case you found that 0 = 4 as a result! For this reason it is best to avoid dividing by terms that may be zero - factorise instead.

Quote:
 Originally Posted by hyperbola $\displaystyle 2(x-1) = x$
This time you didn't come out with nonsense because there is a solution under which $x-1 \ne 0$, but you did lose four (repeated) solutions to the equation.

Again, factorise it rather than divide.

Last edited by skipjack; February 20th, 2015 at 07:24 AM.

February 20th, 2015, 03:29 AM   #8
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Quote:
 Originally Posted by hyperbola Just a quick question: $\displaystyle (x-3)^3 - (x+1)(x-3)^2 = 0$ can the above be solved as such: $\displaystyle (x-3)^3 - (x+1)(x-3)^2 = 0$ $\displaystyle (x-3)^3 = (x+1)(x-3)^2$ $\displaystyle (x-3)^3/(x-3)^2 = (x+1)$
IF x- 3 is not 0, which is the same as x is not 3, we can do this.

Quote:
 $\displaystyle (x-3) = (x+1)$
Obviously there is no value of x that satisfies this so "x = 3" is the only solution.

Quote:
 if so, then could the following be solved the same way? $\displaystyle 2(x-1)^5 - x(x-1)^4 = 0$ $\displaystyle 2(x-1)^5 = x(x-1)^4$ $\displaystyle 2(x-1)^5/(x-1)^4 = x$
Again, IF x- 1 is not 0, which is the same as "x is not equal to 1" then we can divide by it.

Quote:
 $\displaystyle 2(x-1) = x$
This is the same as 2x- 2 = x or x = 2. So the solutions to this equation are x = 1 and x = 2.

Quote:
 Thank you in advance for the feedback.

Last edited by skipjack; February 20th, 2015 at 07:26 AM.

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