# Completing the square when the coeffecient of x^2 is not 1

Discussion in 'Math' started by amilton542, Apr 12, 2011.

1. ### amilton542 Thread Starter Active Member

Nov 13, 2010
496
64
I'm having trouble reaching a given answer

2x^2 + 8x - 25 = 0

Answer: -2 + or - sqr/rt 33/2
My workings out:

2(x^2 + 4x - 25/2) = 0
2(x + 4)^2 - 16 - 25/2

-16 - 25/2 = -32/2 - 25/2 = - 57/2

2(x+4)^2 - 57/2 = 0
2x = -4 + or - sqr/rt 57/2

This is where im stuck. I dont no what to do with the coeffeicient, If I divide the 2 to eleminate it I have to divide everything on the right by 2 and that won't give me the answer

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
First of all, are you sure your answer is correct? From the 2nd grade root formula I get:
$
a=2,b=8,c=-25\\
\Delta=b^2-4\cdot a \cdot c = 64+ 4 \cdot 2 \cdot 25=264\\
x_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2 \cdot a}=\frac{-8 \pm \sqrt{264}}{4} \simeq -6.025\ and\ 2.0625$

Other than that, you correctly factored out 2 on your first step. Just take the expression in the parenthesis and equate it with 0. You now have another, equivalent equation with 1 as the coefficient of the 2nd grade term.

Last edited: Apr 12, 2011
3. ### Papabravo Expert

Feb 24, 2006
11,163
2,186
The use of the phrase "2nd grade" is kind of amusing. Your meaning is the same as the word "quadratic" as in "quadratic formula" for finding roots or "quadratic term" referring to the term with x^2.

In the US public school system "2nd Grade" refers to the second year of elementary education where children are usually about 7 years of age. I know the trend is to push mathematical concepts into lower grades but I'm pretty sure we are not there yet.

That was a good one.

Last edited: Apr 12, 2011
4. ### amilton542 Thread Starter Active Member

Nov 13, 2010
496
64
http://www.mathtutor.ac.uk/algebra/completingthesquare/text Page 8, excercise 4, question f. The answers for the excercises are below. They've used a different method to the quadratic formula

5. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
@ Papabravo
It was a direct translation from Greek. I 'll remember it next time.

@amilton542
I didn't understand that "+ or -" was supposed to mean $\pm$. Yes, that solution is correct.
I know the way you are trying to solve the problem, and if you can do it for a=1 you can do it for any a. Just factor a out and solve the new quadratic equation that is in the parenthesis and has a=1.

6. ### amilton542 Thread Starter Active Member

Nov 13, 2010
496
64
I do not know how the latex reference works

7. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,699
912
If you want to do it by completing the square rather than by using the quadratic formula, here is how:

1) Factor out the coefficient of x$^{2}$ to give 2(x$^{2}$ +4x -12.5) = 0

2) Divide both sides by 2 to give: x$^{2}$ +4x -12.5 =0

3) Solve for y in this equation -12.5 +y =4; y=16.5

4) Now add 16.5 to both sides to get: x$^{2}$ +4x +4 = 16.5

5) (x+2)$^{2}$ = 16.5

6) x+2 = (16.5)$^{1/2}$

John

8. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,699
912
There is a small, but perhaps confusing typo.

$\Delta = b^2 -4\cdot a\cdot c$

John

9. ### Georacer Moderator

Nov 25, 2009
5,151
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Right, typo corrected. Thanks for the correction.

10. ### Papabravo Expert

Feb 24, 2006
11,163
2,186
I know that Greek is your first language and I was not being critical. Translation quite often produces amusing results given the general ability to take account of idiomatic expressions and forms especially in a technical context. One evening while in Italy on assignment the bulb in the reading light in my room went out. The desk clerk was completely nonplussed when I informed him in halting Italian that "the streetlight in my room was not working". My accent was certainly not Piemotese which probably gave him an additional clue.

11. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Don't worry, no offence was taken whatsoever. Actually, I 'm a very hard person to offend.
I 've had my moments with foreign languages too, and it seems I still do.