Quadratic Formula and 4x^2 - 12x + 3 > 0

[Monkeyseat]

Hi,

I used the quadratic formula to find the roots of this curve equation:

4x^2 - 12x + 3 > 0

Anyway, it's a bit fiddly to type out (don't know how to use that TeX thing) so baisically after putting a, b and c in the quadratic formula and figuring it out I got:

I simplified that down to:

The book says the final answer is (which I can't get to):

I know this is the same, but using the quadratic formula to start with, how can I cancel down to that? I can get that answer using completing the square and I was told the quadratic formula comes out with exactly the same answers but I can't get the same answer (that the book gives) on this question with it.

Baisically, using the quadratic formula, how can I get or cancel down to the final answer the book gave?

Sorry wasn't sure whether this was to go under fractions or algebra.

Thanks.

 

[jwpaine]

because \(\displaystyle \L \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{3}sqrt{2}}{sqrt{2}sqrt{2}} = \frac{\sqrt{6}}{2}\)

Your solution has been simplified more - it's better IMHO

 

[o_O]

Your answer is correct. Just a bit of work to get it to the book's answer (which are both equivalent if you plug it in a calculator):

\(\displaystyle \frac{3 \pm \sqrt{6}}{2}\)

\(\displaystyle = \frac{3}{2} \pm \frac{\sqrt{6}}{2}\) (Split up the fraction)

Now imagine 2 as \(\displaystyle \sqrt{4}\):

\(\displaystyle = \frac{3}{2} \pm \frac{\sqrt{6}}{\sqrt{4}}\)

Now, here's a property of square roots that would help you simplify your answer to the book's:
\(\displaystyle \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\)

Edit: Oh. Beaten by John :wink:

 

[Monkeyseat]

because \(\displaystyle \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{3}\sqrt{2}}{\sqrt{2}\sqrt{2}} = \frac{\sqrt{6}}{2}\)

Your solution has been simplified more - it's better IMHO

Thanks to both of you for helping.

So really, it wouldn't make sense to work back that way to the books answer? That is:

\(\displaystyle \frac{\sqrt{6}}{2}\, =\, \frac{\sqrt{3}\sqrt{2}}{\sqrt{2}\sqrt{2}}\, =\, \frac{\sqrt{3}}{\sqrt{2}}\, =\, \sqrt{\frac{3}{2}}\)

There's no real reason to split the bottom up into \(\displaystyle \sqrt{2}\sqrt{2}\) is there (because it's already pretty simple)? The root-2's cancel in the numerator and denominator, right?

\(\displaystyle = \frac{3}{2} \pm \frac{\sqrt{6}}{\sqrt{4}}\)

Now, here's a property of square roots that would help you simplify your answer to the book's:

\(\displaystyle \frac{\sqrt{a}}{\sqrt{b}}\, =\, \sqrt{\frac{a}{b}}\)

So: \(\displaystyle \frac{\sqrt{6}}{\sqrt{4}}\) goes to \(\displaystyle \sqrt{\frac{6}{4}\) and cancels to \(\displaystyle \sqrt{\frac{3}{2}}\) because both divide by 3? Like above, would it still be best to keep the bottom as 2 instead of route 4 because it's "nicer" and rational?

What I'd like to know is please could you possibly show the correct way to get the book answer from here:

It would make more sense than simplifying it like I did, then "unsimplifying" it to the book answer. I just want to be able to show both ways.

Many thanks. Any reply is appreciated.

 

[o_O]

I would just leave it as the way you have it. Look at my post again. You just carry on from there:

\(\displaystyle \L \frac{3}{2} \pm \frac{\sqrt{6}}{\sqrt{4}}\)

\(\displaystyle \L = \frac{3}{2} \pm \sqrt{\frac{6}{4}}\)

\(\displaystyle \L = \frac{3}{2} \pm \sqrt{\frac{3}{2}}\)

No point in doing this really. Your answer is perfectly fine.

 

[jwpaine]

So really, it wouldn't make sense to work back that way to the books answer?

Both are acceptable solutions, but I wouldn't wast your time getting your worked solution into the same form as the book's: it would be redundant.

Good Job.

Cheers,
John

 

[Monkeyseat]

Okay, thanks guys, I understand what you're both saying.

I know it doesn't really matter like you said but I'm just wondering if there is a way that gets there quicker or doesn't have to be simplified, to be "unsimplified" like I did. Or would it always be like that?

Just out of interest though, would someone be able to show how to turn this:

Into this:

In a way that possibly does it a bit better or quicker unlike what I did. Like show a way that makes more "mathmatical" sense (although it probably doesn't make common sense, I know there's no real reason but I'd just like to know really, heh sorry) - not like unsimplifying my answer to get there, just getting directly there from (12 +-sqr. 96)/8. Is there no easier way?

The book probably got this answer by completing the square although on other questions it leaves its answers like I did originally - maybe the numbers are easier in this one.

I know my answer is probably better, it's just bugging me a bit and I'd like to know if possible.

Many thanks. Any help is always appreciated. Baisically if someone could just work through that it would be great. Sorry to keep going on about it, just that last thing.

Cheers.

 

[o_O]

What do you mean? You're not really "unsimplifying" it, you're just finding an equivalent expression that requires work. No matter what you do, you'll end up splitting your original expression into two fractions if you want to end up with your book's answer:

\(\displaystyle \frac{12 \pm \sqrt{96}}{8}\)

\(\displaystyle =\frac{12}{8} \pm \frac{\sqrt{96}}{8}\)

\(\displaystyle =\frac{3}{2} \pm \frac{\sqrt{16 \cdot 6}}{8}\)

\(\displaystyle = \frac{3}{2} \pm \frac{\;4\sqrt{6}\;}{\;4 \cdot 2\;}\)

\(\displaystyle = \frac{3}{2} \pm \frac{\sqrt{6}}{\sqrt{4}}\) . . .\(\displaystyle \leftarrow\) (same thing again as the other post)

\(\displaystyle = \frac{3}{2} \pm \sqrt{\frac{6}{4}\)

\(\displaystyle = \frac{3}{2} \pm \sqrt{\frac{3}{2}\)

 

[Monkeyseat]

Sorry for slow reply, just to say thanks for the help.