February 16th, 2015, 11:51 PM  #1 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty  Quick question.
Just a quick question: $\displaystyle (x3)^3  (x+1)(x3)^2 = 0$ can the above be solved as such: $\displaystyle (x3)^3  (x+1)(x3)^2 = 0$ $\displaystyle (x3)^3 = (x+1)(x3)^2$ $\displaystyle (x3)^3/(x3)^2 = (x+1)$ $\displaystyle (x3) = (x+1)$ if so, then could the following be solved the same way? $\displaystyle 2(x1)^5  x(x1)^4 = 0$ $\displaystyle 2(x1)^5 = x(x1)^4$ $\displaystyle 2(x1)^5/(x1)^4 = x$ $\displaystyle 2(x1) = x$ Thank you in advance for the feedback. 
February 16th, 2015, 11:55 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs 
For the first problem: $\displaystyle (x3)^3(x+1)(x3)^2= 0$ I would factor: $\displaystyle (x3)^2\left((x3)(x+1)\right)= 0$ Collect like terms: $\displaystyle 4(x3)^2= 0$ Hence: $\displaystyle x=3$ Can you apply the same method to the second problem? 
February 17th, 2015, 12:30 AM  #3 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty 
$\displaystyle 2(x1)^5  x(x1)^4$ common factor is $\displaystyle (x1)^4$ $\displaystyle (x1)^4(2(x1)x) = 0$ $\displaystyle (x1)^4(2x2x) = 0$ $\displaystyle (x1)^4(x2) = 0$ $\displaystyle (x1)^4 = 0, x=1$ $\displaystyle (x2) = 0, x=2$ correct? Last edited by hyperbola; February 17th, 2015 at 12:32 AM. 
February 17th, 2015, 01:48 AM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs 
Yes...perfect! 
February 17th, 2015, 04:44 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 05:38 AM. 
February 17th, 2015, 10:14 AM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 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: $\displaystyle x=\frac{6\pm\sqrt{30}}{2}$ 
February 17th, 2015, 12:29 PM  #7  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,094 Thanks: 2360 Math Focus: Mainly analysis and algebra  Quote:
This time you didn't come out with nonsense because there is a solution under which $x1 \ne 0$, but you did lose four (repeated) solutions to the equation. Again, factorise it rather than divide. Last edited by skipjack; February 20th, 2015 at 08:24 AM.  
February 20th, 2015, 04:29 AM  #8  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,879 Thanks: 766  Quote:
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Last edited by skipjack; February 20th, 2015 at 08:26 AM.  

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