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February 16th, 2015, 11:51 PM   #1
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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.
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February 16th, 2015, 11:55 PM   #2
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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?
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February 17th, 2015, 12:30 AM   #3
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$\displaystyle 2(x-1)^5 - x(x-1)^4$

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?
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Last edited by hyperbola; February 17th, 2015 at 12:32 AM.
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February 17th, 2015, 01:48 AM   #4
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Yes...perfect!
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February 17th, 2015, 04:44 AM   #5
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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.
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February 17th, 2015, 10:14 AM   #6
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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:

$\displaystyle x=\frac{-6\pm\sqrt{30}}{2}$
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February 17th, 2015, 12:29 PM   #7
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Quote:
Originally Posted by hyperbola View Post
$\displaystyle (x-3)^3 = (x+1)(x-3)^2$
$\displaystyle (x-3)^3/(x-3)^2 = (x+1)$
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 View Post
$\displaystyle 2(x-1) = x$
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.
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Last edited by skipjack; February 20th, 2015 at 08:24 AM.
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February 20th, 2015, 04:29 AM   #8
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Quote:
Originally Posted by hyperbola View Post
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 08:26 AM.
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