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C1 - Completing The Square

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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Reply 1

KINGYusuf

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 :smile:

Reply 2

AdeptDz

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 :smile:

I know how to and i've done it, i just want to make sure im right because the answers are weird.
Thanks anyway.

Reply 3

DamnDaniel2

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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Reply 4

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)

Reply 5

DamnDaniel2

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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Reply 6

AdeptDz

Original post by DamnDaniel2

Ok lol confusing but i'll try and work it out

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lool i know sorry
Thanks

Reply 7

DamnDaniel2

Yh I think I got what you got if I didn't misunderstand what u wrote ahahaha

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Reply 8

AdeptDz

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

Reply 9

DamnDaniel2

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 :smile:

which subs u doing?

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Reply 10

B_9710

\displaystyle x=\frac{1}{10} (-3\pm \sqrt{29} )

Reply 11

AdeptDz

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} )

Reply 12

Muttley79

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.

Reply 13

AdeptDz

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.

[br]5x^2 + 3x - 1 = 0[br]5(x^2 + \frac{3x}{5})-1 = 0[br]5[(x+{3}/{10})^2 - \frac{9}{100}]-1 = 0[br]5(x+3/10)^2 - \frac{9}{20}-1 = 0[br]5(x+3/10)^2 - \frac{29}{20} = 0[br]5(x+3/10)^2 = \frac{29}{20} [br]5(x+3/10)= \pm{\sqrt{\frac{29}{20}}} [b]error is here[/b][br]x+3/10= \pm{\sqrt{\frac{29/20}{5}}}[br]x = \pm{\sqrt{\frac{29/20}{5}}} - \frac{3}{10}[br]

(edited 7 years ago)

Reply 14

AdeptDz

give me a second to reformat tht

Reply 15

AdeptDz

There.
May someone help me?
Sorry its a bit messy

Reply 16

AdeptDz

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