My Math Forum Completing the Square question.

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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
Math Team

Joined: Mar 2012
From: India, West Bengal

Posts: 3,871
Thanks: 86

Math Focus: Number Theory
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
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

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: 7,968 Thanks: 1152 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
Newbie

Joined: Jan 2014

Posts: 4
Thanks: 0

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
Member

Joined: Dec 2013

Posts: 82
Thanks: 0

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
Newbie

Joined: Jan 2014

Posts: 4
Thanks: 0

Re: Completing the Square question.

Quote:
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: 7,968 Thanks: 1152 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
Member

Joined: Dec 2013

Posts: 82
Thanks: 0

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.

 Tags completing, question, square

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post link107 Algebra 2 September 8th, 2012 02:42 PM bearsfan092 Algebra 3 January 24th, 2010 05:09 PM -DQ- Algebra 3 September 10th, 2009 05:51 PM PistolPete Algebra 15 May 20th, 2009 06:44 PM milly2012 Abstract Algebra 2 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top