C1 - Completing The Square Watch

AdeptDz
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#1
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#1
I just want to check my answer.
Could someone solve for x by using completing the square on this:
 5x^2 + 3x - 1
Thanks
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KINGYusuf
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(Original post by AdeptDz)
I just want to check my answer.
Could someone solve for x by using completing the square on this:
 5x^2 + 3x - 1
Thanks
I would advise you watch ExamSolutions tutorial and then attempt it yourself. It's not that difficult and once you watch you won't forget !
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AdeptDz
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(Original post by KINGYusuf)
I would advise you watch ExamSolutions tutorial and then attempt it yourself. It's not that difficult and once you watch you won't forget !
I know how to and i've done it, i just want to make sure im right because the answers are weird.
Thanks anyway.
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DamnDaniel2
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(Original post by AdeptDz)
I know how to and i've done it, i just want to make sure im right because the answers are weird.
Thanks anyway.
What did you get?


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AdeptDz
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i don't even know how to right it in latex so ill try type it

x = ((plus or minus)(squareroot of 29 over 20) minus (3 over 2)) all over (5)
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DamnDaniel2
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(Original post by AdeptDz)
i don't even know how to right it in latex so ill try type it

x = ((plus or minus)(squareroot of 29 over 20) minus (3 over 2)) all over (5)
Ok lol confusing but i'll try and work it out


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AdeptDz
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(Original post by DamnDaniel2)
Ok lol confusing but i'll try and work it out


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lool i know sorry
Thanks
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DamnDaniel2
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Yh I think I got what you got if I didn't misunderstand what u wrote ahahaha


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AdeptDz
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(Original post by DamnDaniel2)
Yh I think I got what you got if I didn't misunderstand what u wrote ahahaha


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cool thanks, ill double check anyway by doing the quadratic equation, thats the second question anyway
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DamnDaniel2
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(Original post by AdeptDz)
cool thanks, ill double check anyway by doing the quadratic equation, thats the second question anyway
Ahaha oh ok! Just saying im also in year 12 so my maths isn't amazing which subs u doing?


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B_9710
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 \displaystyle x=\frac{1}{10} (-3\pm \sqrt{29} ) .
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AdeptDz
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yh thanks, i see where i went wrong so ill try again

(Original post by B_9710)
 \displaystyle x=\frac{1}{10} (-3\pm \sqrt{29} ) .
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Muttley79
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(Original post by AdeptDz)
I just want to check my answer.
Could someone solve for x by using completing the square on this:
 5x^2 + 3x - 1
Thanks
It's a good idea to post your working if you want someone to help in future.
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AdeptDz
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Woah can someone explain what I'm doing wrong? this approach worked on all the other questions but this one, or maybe im doing the algebra wrong or something.



5x^2 + 3x - 1 = 0

5(x^2 + \frac{3x}{5})-1 = 0

5[(x+{3}/{10})^2 - \frac{9}{100}]-1 = 0

5(x+3/10)^2 - \frac{9}{20}-1 = 0

5(x+3/10)^2 - \frac{29}{20} = 0

5(x+3/10)^2 = \frac{29}{20} 

5(x+3/10)= \pm{\sqrt{\frac{29}{20}}} <b>error is here</b>

x+3/10= \pm{\sqrt{\frac{29/20}{5}}}

x = \pm{\sqrt{\frac{29/20}{5}}} - \frac{3}{10}
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AdeptDz
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give me a second to reformat tht
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AdeptDz
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There.
May someone help me?
Sorry its a bit messy
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AdeptDz
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Oh never mind i see my error
It is when i square root both sides right? i don;t square root the 5 aswell, so it wud be better to divide by 5 then square root
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Naruke
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(Original post by AdeptDz)


5[(x+{3}/{10})^2 - \frac{9}{100}]-1 = 0

5(x+3/10)^2 - \frac{9}{20}-1 = 0
What did u do here
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MajorFader
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Also, you didn't take 5 out as a factor for the C value.
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AdeptDz
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(Original post by Naruke)
What did u do here
i mulitplied everything in square bracekt by 5
But yeh i see my mistake.. should divide by 5 then square root to get an answer of:



{-3}/{10} \pm \sqrt{\frac{29}{100}}
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