Finding roots - C1

I have this question, but I just can't get the right answer. I dunno if I'm doing the wrong method or just stupid mistakes:

Find the roots of 2x^2 - 4x - 3 = 0

I use completing the square method, by taking the coefficient of x out first:

2(x^2 - 4x -3) = 0

Then factorise that:

2((x - 2)^2 - 7) = 0

Then take the 7 over:

2(x - 2)^2 = 7

And then I'm not really sure what to do so I divide both sides by 2:

(x - 2)^2 = 7/2

Then root both sides and then add 2 to get the final answer of 2+/- (rt) 7/2.

However, the answer is apparently 1 +/- (rt) 10/2.

HELP!!??

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Reply 1

gdunne42

Original post by Conal09

2x^2 - 4x - 3 does not = 2(x^2 - 4x -3)

multiply it out and you get 2x^2 - 8x - 6

Reply 2

little_wizard123

The very first line is wrong; try multiply it out to see why.

Reply 3

Conal09

Even so, I still don't get the right answer. Now I get 1 +/- (rt) 1.25

Reply 4

EEngWillow

Original post by Conal09

Even so, I still don't get the right answer. Now I get 1 +/- (rt) 1.25

Post your new working?

Reply 5

Freakonomics123

why dont u just put it in quadratic equation?

Reply 6

Conal09

Original post by EEngWillow

Post your new working?

2(x^2 - 2x - 3/2) = 0

2((x - 1)^2 - 1 - 3/2 = 0

2(x - 1)^2 = 2 1/2

(x - 1)^2 = 1 1/4

x - 1 = +/- (rt) 1 1/4

x = 1 +/- (rt) 1 1/4

Bound to be something wrong in there.

Reply 7

EEngWillow

Original post by Conal09

2(x^2 - 2x - 3/2) = 0

2((x - 1)^2 - 1 - 3/2 = 0

2(x - 1)^2 = 2 1/2

(x - 1)^2 = 1 1/4

x - 1 = +/- (rt) 1 1/4

x = 1 +/- (rt) 1 1/4

Bound to be something wrong in there.

Is that (2 and a half) or 2(1/2)?
If the latter, there's your problem (What's 2(1 + 3/2). Otherwise your problem is the next line.

Reply 8

gdunne42

Original post by Conal09

2(x^2 - 2x - 3/2) = 0

2((x - 1)^2 - 1 - 3/2 = 0

2(x - 1)^2 = 2 1/2

(x - 1)^2 = 1 1/4

x - 1 = +/- (rt) 1 1/4

x = 1 +/- (rt) 1 1/4

Bound to be something wrong in there.

2((x - 1)^2 - 1 - 3/2) = 0 missing bracket
so
(x - 1)^2 - 1 - 3/2 = 0
then
(x - 1)^2 = 5/2 (I recommend you leave it as a fraction)

Reply 9

Conal09

Original post by EEngWillow

Is that (2 and a half) or 2(1/2)?
If the latter, there's your problem (What's 2(1 + 3/2). Otherwise your problem is the next line.

No, it's 2 and a half.

Reply 10

Conal09

Original post by gdunne42

2((x - 1)^2 - 1 - 3/2) = 0 missing bracket
so
(x - 1)^2 - 1 - 3/2 = 0
then
(x - 1)^2 = 5/2 (I recommend you leave it as a fraction)

And this still gives me the same answer anyway since 5/2 divided by 2 = 1 1/4

Reply 11

MostCompetitive

If you can't complete the square, use the formula. So the roots are given by

\frac{4+\sqrt{16-(-24)}}4

and

\frac{4-\sqrt{16-(-24)}}4

Reply 12

Conal09

Original post by MostCompetitive

If you can't complete the square, use the formula. So the roots are given by

\frac{4+\sqrt{16-(-24)}}4

and

\frac{4-\sqrt{16-(-24)}}4

okay, I got it :smile:

thanks. I didn't really give that method a thought cos you usually have to use calculators :smile:

Reply 13

bb193

There seems to be a problem with taking the 2 outside the brackets. I'd try just taking the first line and dividing the whole lot by 2, so you just get an x2 term, the 2 disappears and it can cause no further problems.

Reply 14

MostCompetitive

Original post by Conal09

okay, I got it :smile:

thanks. I didn't really give that method a thought cos you usually have to use calculators :smile:

You only have to use a calculator when you're not going to leave the roots in surd form. You're welcome.

(edited 13 years ago)

Reply 15

EEngWillow

Original post by Conal09

okay, I got it :smile:

thanks. I didn't really give that method a thought cos you usually have to use calculators :smile:

Since you have the answer now I'm just going to post up the CTS solution.

2x^2 - 4x - 3 = 0[br][br]2(x^2 - 2x - \frac{3}{2}) = 0[br][br]2[(x-1)^2 - 1 - \frac{3}{2}] = 0[br][br]2(x-1)^2 = 5[br][br](x-1)^2 = \frac{5}{2}[br][br]x-1 = \sqrt{\frac{5}{2}}[br][br]x = \pm\sqrt{\frac{5}{2}} + 1 = 1 \pm \frac{\sqrt{10}}{2}[br]

Reply 16

bb193

I hope this isn't taken the wrong way.

Solving quadratics isn't really too difficult, and this one should be fairly straightforward to solve both ways pretty quickly. It isn't that one way is better than the other - you need to be equally happy with both.

It is worth becoming familiar with the techniques, so that when you see a quadratic in the exam, your little heart skips a beat as you know you have got 100% on that question. Honestly, you can get there without too much effort.

So if I were you I'd take 20 questions like that and solve each of them both ways. If towards the end of the 20, you are getting them all right in less than a couple of minutes each, then stop, pat yourself on the back and tell yourself you have got it. If not - try another 10.

Reply 17

lukas1051

Original post by Conal09

I'm doing C1 also, when I get a completing the square question with a coefficient in front of x^2 I find it much less confusing to divide the whole thing through by 2 rather than taking 2 outside of the brackets like you did. You get some pretty nasty fractions, but it all works out in the end.

Alternatively, you could just use the quadratic formula if your surd arithmetic is good.

Reply 18

Conal09

Original post by EEngWillow

Since you have the answer now I'm just going to post up the CTS solution.

2x^2 - 4x - 3 = 0[br][br]2(x^2 - 2x - \frac{3}{2}) = 0[br][br][b]2[(x-1)^2 - 1 - \frac{3}{2}] = 0[br][br]2(x-1)^2 = 5[/b][br][br](x-1)^2 = \frac{5}{2}[br][br]x-1 = \sqrt{\frac{5}{2}}[br][br]x = \pm\sqrt{\frac{5}{2}} + 1 = 1 \pm \frac{\sqrt{10}}{2}[br]

Can you just confirm this for me? So you have -2 1/2 and then, to get -5, you multiply through by 2? And in doing so you get 2(x-1)^2? That's where I went wrong in that method :colondollar: