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I've been searching all over how to handle this issue I'm sitting with. Hope someone can help me out :)

I have to convert sentence (1) to (6) by removing the quadratic term in (1) with substitution (x = (t - a/3). I feel like I'm doing something wrong when calculation the squared parenthesis when there's a number in front of it as I'm suppose to remove the quadratic term? Can someone please take a look at the calculations and correct me if I'm wrong?

Thanks in advance!

image over calculations so far

eagle
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1 Answers1

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The substitution $t-\frac a3$ is for a cubic equation of the form $x^3 + ax^2 + bx + c = 0.$

In order to make that formula apply to your particular equation, $x^3 + 1.5x^2 -2.5x + 2 = 0,$ all you need to do (and indeed what you must do) is to set $a=1.5,$ $b=-2.5,$ and $c=2.$

Having done that, you may (and in this case should) make the substitution $a = 1.5$ everywhere where the symbol $a$ occurs. If you do this, your $x^2$ terms will cancel.

You have a transcription error in the coefficient of your linear term, by the way; it was $-2.5$ initially but is later written $+2.5.$

David K
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  • Thanks for the reply David K.

    Does this look right? http://i.imgur.com/c8PGmCX.png

     – eagle Feb 12 '17 at 21:57
  • Yes, that looks good. If I were doing it I would probably convert $\frac a3$ from $\frac{1.5}{3}$ to $0.5$ or $\frac12$ right away, but your way also works. Notice that if you add the right sides of equations 3 and 4 now, the $t^2$ terms will cancel, which is the result you should get from this method. – David K Feb 13 '17 at 02:00
  • Hey again David. Can you take a look at the result and confirm if this is correct?

    http://i.imgur.com/3iZugzn.png

     – eagle Feb 13 '17 at 10:55
  • It looks much better. The only thing that didn't look right is the $3.25t$ in formula (7); it looks like you did not account for the term $-1.5t.$ – David K Feb 16 '17 at 22:28