Factorise using completed square form?

**Miasmagasma** · December 7th 2009, 02:33 PM

use the completed square form to factorise the following expression.

14 + 45x - 14x

i know how to solve it with trial and error, but when i put the numbers into completed square form, it is far far too messy to be right. i'm doing something drastically wrong.

would love to be walked through this one.

cheers.

**RRH** · December 7th 2009, 02:39 PM

Originally Posted by Miasmagasma

use the completed square form to factorise the following expression.

14 + 45x - 14x

i know how to solve it with trial and error, but when i put the numbers into completed square form, it is far far too messy to be right. i'm doing something drastically wrong.

would love to be walked through this one.

cheers.

Could you check the equation please .. One of the variables has to be $x^2$

**e^(i*pi)** · December 7th 2009, 02:43 PM

Originally Posted by Miasmagasma

use the completed square form to factorise the following expression.

14 + 45x - 14x

i know how to solve it with trial and error, but when i put the numbers into completed square form, it is far far too messy to be right. i'm doing something drastically wrong.

would love to be walked through this one.

cheers.

Methinks there is a typo as there is no x^2 to be seen which is required

**Miasmagasma** · December 7th 2009, 02:51 PM

ahhh sorry

it's 14x^2

**Miasmagasma** · December 7th 2009, 03:26 PM

i am taking a levels 10 years since i finished gcse and i think it's becoming a huge hindrance ( i'm working by myself from home ).

i know i'm making small mistakes which are constantly making hard work of things i'm perfectly capable of doing.

i've just done this

4 + 6x - x^2

-->

4 + [( 3 - x )^2 - 9]

-->

( 3 - X )^2 -5

lowest point is -5 when x = 3

the right answer is 13 when x = 3 so i'm clearly making a stupid mistake, as i can see that 9 + 4 is 13 yet can't see how to end up with the right answer. it's getting me down now. anyone know what i'm doing wrong and how i can brush up on it?

**RRH** · December 7th 2009, 03:27 PM

Originally Posted by Miasmagasma

ahhh sorry

it's 14x^2

I like to start by rearrange the equation into standard form

$ -14x^2+45x+14=0 $

First divide all variables by -14 to make the "a" variable equal to 1; therefore, it should now look like this

$ x^2-\frac{45}{14}x-1=0 $

$ \frac{\frac{45}{14}}{2} $

$ \frac{45}{14}*\frac{1}{2} $

$ \frac{45}{28} $

$ (\frac{45}{28})^2 $

$ \frac{2025}{784} $

$ x^2-\frac{45}{14}x+\frac{2025}{784}=1+\frac{2025}{784} $

$ (x-\frac{45}{28})^2=\frac{2809}{784} $

$ \sqrt [2] {(x-\frac{45}{28})^2}=\sqrt [2] {\frac{2809}{784}} $

$ x-\frac{45}{28}=\frac{53}{28} and x-\frac{45}{28}=-\frac{53}{28} $

x equals $\frac{7}{2}$ and $-\frac{2}{7}$

**RRH** · December 7th 2009, 03:40 PM

Originally Posted by Miasmagasma

i am taking a levels 10 years since i finished gcse and i think it's becoming a huge hindrance ( i'm working by myself from home ).

i know i'm making small mistakes which are constantly making hard work of things i'm perfectly capable of doing.

i've just done this

4 + 6x - x^2

-->

4 + [( 3 - x )^2 - 9]

-->

( 3 - X )^2 -5

lowest point is -5 when x = 3

the right answer is 13 when x = 3 so i'm clearly making a stupid mistake, as i can see that 9 + 4 is 13 yet can't see how to end up with the right answer. it's getting me down now. anyone know what i'm doing wrong and how i can brush up on it?

$ -x^2+6x+4=0 $

$ -1(x^2-6x-4)=0 $

$ -1(x^2-6x+9-9-4)=0 $

$ -1(x^2-6x+9)+13=0 $

$ -(x-3)^2+13 $

**Miasmagasma** · December 7th 2009, 04:01 PM

Originally Posted by RRH

$ -x^2+6x+4=0 $

$ -1(x^2-6x-4)=0 $

$ -1(x^2-6x+9-9-4)=0 $

$ -1(x^2-6x+9)+13=0 $

$ -(x-3)^2+13 $

hmmm the '+9 -9 -4' has thrown me. i have no idea where it came from?

**RRH** · December 7th 2009, 04:23 PM

Originally Posted by Miasmagasma

hmmm the '+9 -9 -4' has thrown me. i have no idea where it came from?

original equation

$-x^2+6x+4=0$

factor out negative one

$-1(x^2-6x-4)=0$

Now we divide the -6 by 2 and square the result, which is 9; therefore, we will have to add +9 and -9 to the equation within the bracket.

$-1(x^2-6x+9-9-4)=0$

The +9 and the -9 come from the process of completing the square the -4 is from the +4 from the original equation with the -1 factored out

I hope that helps ... If it does not keep asking questions and we will take care of you

**Miasmagasma** · December 7th 2009, 05:47 PM

really appreciate the help, friend.

i don't recall doing anything like what you have just shown me.

what is this halving and squaring of 6 ( which i know as 'b' ) based on? what equation is it based on?

i have only come across [ a(x + 1/2b )^2 - 1/4b^2 ] + c

**RRH** · December 7th 2009, 06:36 PM

Originally Posted by Miasmagasma

really appreciate the help, friend.

i don't recall doing anything like what you have just shown me.

what is this halving and squaring of 6 ( which i know as 'b' ) based on? what equation is it based on?

i have only come across [ a(x + 1/2b )^2 - 1/4b^2 ] + c

Here is a website with a visual representation of completing the square

How to Complete the Square

look at the green box ... it shows all the steps

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