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Completing the square

Hi, can someone show me a step by step method for completing the square for

9x^2 - 12x - 2 = 0?

What I did:

3(3x - 4)^2 - 2 = 0
3(3x - 2)^2 = 6
3(3x - 2) = √6
3x - 2 = √6 / 3
3x = 2 + √6 / 3 (all over 3)
Reply 1
Avatar for JezDayy
JezDayy
OP
11
I forgot to say I need to solve for x. The answer is meant to be
Screen Shot 2016-05-07 at 20.19.35.png
and what I got in the first post is wrong because I have 3x instead of x =.
This is the question
Attachment not found

This is the whole mark scheme for part (c)

Screen Shot 2016-05-07 at 20.20.37.png
Thank you.
(edited 7 years ago)
Screen Shot 2016-05-07 at 20.21.45.png60.5KB
Original post by JezDayy
Hi, can someone show me a step by step method for completing the square for

9x2 - 12x - 2 = 0?

What I did:

3(3x - 4)^2 - 2 = 0
3(3x - 2)^2 = 6
3(3x - 2) = √6
3x - 2 = √6 / 3
3x = 2 + √6 / 3 (all over 3)


write a(bx + c)2 - d = 0

expand this & compare to

9x2 - 12x - 2 = 0

to find a,b,c,d
(edited 7 years ago)
Reply 3
Avatar for JezDayy
JezDayy
OP
11
Original post by the bear
if you multiply out the highlighted expression you get 27x2

Oh yeah I can see that now. Would you take 9 out as a factor before completing the whole thing? I tried that too but got root 22 or something instead of root 6.
Original post by JezDayy
Hi, can someone show me a step by step method for completing the square for

9x^2 - 12x - 2 = 0?

What I did:

3(3x - 4)^2 - 2 = 0
3(3x - 2)^2 = 6
3(3x - 2) = √6
3x - 2 = √6 / 3
3x = 2 + √6 / 3 (all over 3)


you must take out 9 as a common factor first(even if it's not common xD

9(x2−43x−29)=09\left(x^2 - \dfrac{4}{3} x - \dfrac{2}{9}\right)=0

then don't forget since 9 times by everything you do this??

9((x−23)2+43−29)=09\left( \left(x-\dfrac{2}{3}\right)^2 + \dfrac{4}{3} - \dfrac{2}{9}\right)=0

y'know
43−29\dfrac{4}{3} - \dfrac{2}{9}

9((x−23)2+109)=09\left( \left(x-\dfrac{2}{3}\right)^2 + \dfrac{10}{9}\right)=0

then expand outer brackets to get

9(x−23)2+10=09\left(x-\dfrac{2}{3}\right)^2 + 10=0

^^ i think that's right
(edited 7 years ago)
You are supposed to use the quadratic formula for c.
Reply 6
Avatar for JezDayy
JezDayy
OP
11
Original post by thefatone
you must take out 9 as a common factor first(even if it's not common xD9(x2−43x−29)=09\left(x^2 - \dfrac{4}{3} x - \dfrac{2}{9}\right)=0

Oh ok, thank you! Looks nice to work with lol

Original post by greentron6
You are supposed to use the quadratic formula for c.

I know, I realised after looking at the answer, but I was just wondering how to complete the square in general since I couldn't do it and need to work on it.
Original post by JezDayy
Hi, can someone show me a step by step method for completing the square for

9x^2 - 12x - 2 = 0?

What I did:

3(3x - 4)^2 - 2 = 0
3(3x - 2)^2 = 6
3(3x - 2) = √6
3x - 2 = √6 / 3
3x = 2 + √6 / 3 (all over 3)


http://www.examsolutions.net/maths-revision/core-maths/algebra-and-functions/quadratics/completing-the-square/tutorial-1.php

:biggrin:Enjoy!
Original post by JezDayy
Oh ok, thank you! Looks nice to work with lol


I know, I realised after looking at the answer, but I was just wondering how to complete the square in general since I couldn't do it and need to work on it.


i edited it :smile:

this one's a tougher one..... since it's got fractions and you took out a common factor of 9....
Reply 9
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JezDayy
OP
11
Original post by greentron6


haha cheers for the link, but I actually watched this before posting here since I used his method and still got it wrong, and had no idea where I went wrong lol

Original post by thefatone
i edited it :smile:

this one's a tougher one..... since it's got fractions and you took out a common factor of 9....


Thank you! Really appreciate the help. If I knew the quadratic formula was an easier way I would've used it lol, I usually never use it in C1.
Hope this helps! Let me know if any of it doesn't make sense!

Posted from TSR Mobile
1462649909185.jpg52.2KB
Original post by JezDayy
haha cheers for the link, but I actually watched this before posting here since I used his method and still got it wrong, and had no idea where I went wrong lol



Thank you! Really appreciate the help. If I knew the quadratic formula was an easier way I would've used it lol, I usually never use it in C1.


really????

well..... as long as you know how to both ^-^
Original post by JezDayy
Hi, can someone show me a step by step method for completing the square for

9x^2 - 12x - 2 = 0?



see attached file
step by step how to complete...pdf369.3KB
Reply 13
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JezDayy
OP
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Original post by megan-garner6
Hope this helps! Let me know if any of it doesn't make sense!Posted from TSR Mobile
Thank you! Really appreciate this. Is there a rule you need to remember for the root all 2 / 3 becoming root 6 / 3? I don't understand that bit.
Original post by thefatone
really????well..... as long as you know how to both ^-^
Yeah, since it's non-calculator and numbers are usually big or the x values are whole lol

Original post by depymak
see attached file


Thank you!
Original post by JezDayy
Thank you! Really appreciate this. Is there a rule you need to remember for the root all 2 / 3 becoming root 6 / 3? I don't understand that bit.Yeah, since it's non-calculator and numbers are usually big or the x values are whole lol



Thank you!


Haha I used my calculator for that bit! I think you can leave root(2/3) as root(6/9) then just root the 9 to make 3 and leave the root on top of the 6.
Reply 15
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JezDayy
OP
11
Original post by megan-garner6
Haha I used my calculator for that bit! I think you can leave root(2/3) as root(6/9) then just root the 9 to make 3 and leave the root on top of the 6.


Oh lol, I can't use a calculator in that question!! :frown:
I've noticed and made up some kind of rule though lol, if you times the numerator and the denominator of the fraction, put the number you get on the numerator and square root that only, then put that over the original denominator. I need to remember this method haha

Screen Shot 2016-05-07 at 20.54.05.png
Original post by JezDayy
Oh lol, I can't use a calculator in that question!! :frown:
I've noticed and made up some kind of rule though lol, if you times the numerator and the denominator of the fraction, put the number you get on the numerator and square root that only, then put that over the original denominator. I need to remember this method haha

Screen Shot 2016-05-07 at 20.54.05.png


Ooh that's good! I think I'll use that!
Original post by JezDayy
Oh lol, I can't use a calculator in that question!! :frown:
I've noticed and made up some kind of rule though lol, if you times the numerator and the denominator of the fraction, put the number you get on the numerator and square root that only, then put that over the original denominator. I need to remember this method haha

Screen Shot 2016-05-07 at 20.54.05.png


LOL some weird rule eh?
that's called rationalising the denominator, i.e multiply by something to get rid of the surd on the bottom

let's take a look

25=25\sqrt {\dfrac{2}{5}} = \dfrac{\sqrt 2}{\sqrt 5}

multiply top and bottom by 5 \sqrt 5 to get rid of the 5 \sqrt 5on the bottom right?

25×55 \dfrac{\sqrt 2}{ \sqrt 5} \times \dfrac{\sqrt 5}{\sqrt 5}

5×25×5 \dfrac{\sqrt 5 \times \sqrt 2}{\sqrt 5 \times \sqrt 5}

5×25×5 \dfrac{\sqrt {5 \times 2}}{\sqrt {5\times 5}}

everything cancels and works out as

105 \dfrac{\sqrt 10}{5}
(edited 7 years ago)
Reply 18
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JezDayy
OP
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Original post by thefatone
LOL some weird rule eh?
that's called rationalising the denominator, i.e multiply by something to get rid of the surd on the bottom

let's take a look

25=25\sqrt {\dfrac{2}{5}} = \dfrac{\sqrt 2}{\sqrt 5}

multiply top and bottom by 5 \sqrt 5 to get rid of the 5 \sqrt 5on the bottom right?

25×55 \dfrac{\sqrt 2}{ \sqrt 5} \times \dfrac{\sqrt 5}{\sqrt 5}


OH right LOL I know what rationalising the denominator is, I just didn't know you could do it there! I don't think I'm prepared for C1 at all lol
Original post by JezDayy
OH right LOL I know what rationalising the denominator is, I just didn't know you could do it there! I don't think I'm prepared for C1 at all lol


LOL it's ok, i edited it and finished it off xD

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