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 January 8th, 2014, 09:28 AM #1 Newbie   Joined: Jan 2014 Posts: 4 Thanks: 0 Completing the Square question. Hi, I'm currently learning about quadratic equations and 'Completing the Square'. Basically, I need to know what I am doing it correct. I have this equation: x^2 + 5x + 2 = 0 I have worked out that (be it correct or incorrect) that: x = 3.207 or x = 1.792 But when I input one of these numbers into the equation, I don't get an answer that is zero. Have I answered this question correctly? And if so, how can I check I have? If not, what have I done wrong? I can take a photo of my workings out and post them if needs be. Many Thanks
January 8th, 2014, 09:58 AM   #2
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Re: Completing the Square question.

Quote:
 Originally Posted by GuyDoingMathFoiled Have I answered this question correctly?
The equation has two negative roots as easily as can be seen by rule of signs, so definitely no.

January 8th, 2014, 10:01 AM   #3
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Re: Completing the Square question.

Hello, GuyDoingMathFoiled!

Quote:
 $\text{Solve by completing the square: }\:x^2\,+\,5x\,+\,2 \:=\: 0$ I have worked out that (be it correct or incorrect) that:[color=beige] .[/color]$x \,=\,3.207\text{ or }x \,=\, 1.792$ But when I input one of these numbers into the equation, I don't get an answer that is zero. What have I done wrong?

It's hard to see your work from here,
but it looks like you played the $Q\clubsuit$ instead of the $7\spadesuit.$

$\text{W\!e have: }\:x^2\,+\,5x \;=\;-2$

$\text{Then: }\:x^2\,+\,5x\,+\,\frac{25}{4} \;=\;-2\,+\,\frac{25}{4}$

$\;\;\;\left(x\,+\,\frac{5}{2}\right)^2 \;=\;\frac{17}{4}$

$\;\;\;x\,+\,\frac{5}{2}\;=\;\pm\sqrt{\frac{17}{4}} \;=\;\pm\frac{\sqrt{17}}{2}$

$\;\;\;x \;=\;-\frac{5}{2}\,\pm\,\frac{\sqrt{17}}{2} \;=\;\frac{-5\,\pm\,\sqrt{17}}{2} \;=\;\begin{Bmatrix}-0.438447187 \\ \\ \\ -4.561552813\end{Bmatrix}$

 January 9th, 2014, 07:05 AM #4 Newbie   Joined: Jan 2014 Posts: 4 Thanks: 0 Re: Completing the Square question. I don't understand what I'm doing wrong here. I feel so dumb. Here is another equation I've got which for some reason I'm doing wrong. I'll type out my workings out in a hope someone will be able to point me in the right direction. (I don't know how to use the proper mathematical symbols on this forum so bear with me). x^2 -3x - 4 = 0 x^2 -3x = 4 x^2 -3x -(3/2)^2 = 4 -(3/2)^2 (x - 3/2)^2 = 7/4 x - 3/2 = ±?7/4 x = -3/2 ±?7/4 x = -0.177 or -2.823 If the above is too messy to read, please point me towards a site I can convert it into mathematic expressions. Thanks for the help!
 January 9th, 2014, 07:49 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond Re: Completing the Square question. In line 3, try adding (3/2)^2 instead of subtracting it.
January 9th, 2014, 08:15 AM   #6
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Re: Completing the Square question.

Quote:
 Originally Posted by greg1313 In line 3, try adding (3/2)^2 instead of subtracting it.
I thought it had to be -3/2, because to get it you have to divide the -3 by 2.

January 9th, 2014, 08:19 AM   #7
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Re: Completing the Square question.

Quote:
 Originally Posted by GuyDoingMathFoiled I don't understand what I'm doing wrong here. I feel so dumb. Here is another equation I've got which for some reason I'm doing wrong. I'll type out my workings out in a hope someone will be able to point me in the right direction. (I don't know how to use the proper mathematical symbols on this forum so bear with me).
I think what might be misleading you a bit is line 3. You've got:

$x^2 -3x - \left(\dfrac{3}{2}\right)^2= 4 - \left(\dfrac{3}{2}\right)^2$

You can only subtract it from the LHS.

$(x-\dfrac{3}{2})^2 - \left(\dfrac{3}{2}\right)^2= 4$

And then you can move it to the other side by adding it to both sides.

$(x-\dfrac{3}{2})^2= 4 + \left(\dfrac{3}{2}\right)^2$

Personally, I prefer doing it this way than adding it to both sides. It's a question of preference I guess.

Had to edit that a couple of times to make it look correct!

January 9th, 2014, 10:33 AM   #8
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Re: Completing the Square question.

Quote:
Originally Posted by Slinkey
Quote:
 Originally Posted by GuyDoingMathFoiled I don't understand what I'm doing wrong here. I feel so dumb. Here is another equation I've got which for some reason I'm doing wrong. I'll type out my workings out in a hope someone will be able to point me in the right direction. (I don't know how to use the proper mathematical symbols on this forum so bear with me).
I think what might be misleading you a bit is line 3. You've got:

$x^2 -3x - \left(\dfrac{3}{2}\right)^2= 4 - \left(\dfrac{3}{2}\right)^2$

You can only subtract it from the LHS.

$(x-\dfrac{3}{2})^2 - \left(\dfrac{3}{2}\right)^2= 4$

And then you can move it to the other side by adding it to both sides.

$(x-\dfrac{3}{2})^2= 4 + \left(\dfrac{3}{2}\right)^2$

Personally, I prefer doing it this way than adding it to both sides. It's a question of preference I guess.

Had to edit that a couple of times to make it look correct!

Okay, but (and I'm not disputing what you say, just asking to understand better), here:

$(x-\dfrac{3}{2})^2 - \left(\dfrac{3}{2}\right)^2= 4$

You subtract from the left hand side, to make the equation equal, do you not also need to subtract from the right hand side?

 January 9th, 2014, 11:12 AM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond Re: Completing the Square question. $$$x\,-\,\frac32$$^2\,-\,$$\frac32$$^2\,=\,4$ $x^2\,-\,3x\,+\,$$\frac32$$^2\,-\,$$\frac32$$^2\,=\,4$ $x^2\,-\,3x\,=\,4$ See?
January 9th, 2014, 12:26 PM   #10
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Re: Completing the Square question.

Quote:
 Originally Posted by GuyDoingMathFoiled Okay, but (and I'm not disputing what you say, just asking to understand better), here: $(x-\dfrac{3}{2})^2 - \left(\dfrac{3}{2}\right)^2= 4$ You subtract from the left hand side, to make the equation equal, do you not also need to subtract from the right hand side?
Sure, and I am glad that you want to understand. That is the sign of a good student!!

OK, let's just look at the LHS here OK?

We started with

$x^2+3x$

When we square it we get

$(x-\dfrac{3}{2})^2 - \left(\dfrac{3}{2}\right)^2$

Let's take the first term here:

$(x-\dfrac{3}{2})^2$

and expand it.

$(x-\dfrac{3}{2})(x-\dfrac{3}{2})=x^2-3x+\dfrac{9}{4}$

Notice that we have 9/4 too much!!!

That's why we have to subtract it.

Hence:

$x^2+3x=(x-\dfrac{3}{2})^2 - \left(\dfrac{3}{2}\right)^2$

That is why you don't put it on the RHS of the equation as well. The equation is still balanced.

Hope that helps.

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