Factorising quadratics

Original post by RDKGames

To me it seems we have literally written the same thing. Maybe I just haven't explained it "by the book"

sqrt(4) = 2, but x^2 = 4 -> x = +/- sqrt(4) = +/- 2.
If what you said in your post was correct, the +/- wouldn't be needed in the quadratic formula as the square root would already include it.

Reply 22

Original post by HapaxOromenon3

sqrt(4) = 2, but x^2 = 4 -> x = +/- sqrt(4) = +/- 2.
If what you said in your post was correct, the +/- wouldn't be needed in the quadratic formula as the square root would already include it.

Ah I see. Took me a while to notice what the hell you guys were on about. That's not what I was intending in my explanation.

Reply 23

newusername96

Original post by miaofcourse

How do you factorise 2x^2 - 3xI understand you factorise by taking out a common factor, will the factor be X making it x^2 - 2x?? Or am I just being silly, I feel as if it is a simple question but I am going through a mind block! And I'm guessing it's not as simple as x(2x-3)??

since it's in the form of ax^2 + bx and not ax^2 +bx + c (i.e. no c term) simply take out the common factor 'x', so you are correct.

Reply 24

Original post by RDKGames

Ah I see. Took me a while to notice what the hell you guys were on about. That's not what I was intending in my explanation.

For further information see http://www.thestudentroom.co.uk/showthread.php?t=4066671&p=64637655#post64637655

Reply 25

Original post by BasharAssad

since it's in the form of ax^2 + bx and not ax^2 +bx + c (i.e. no c term) simply take out the common factor 'x', so you are correct.

His original query was already answered some time ago, so there's no need for you to post a reply to it at this point.

Reply 26

newusername96

Original post by HapaxOromenon3

His original query was already answered some time ago, so there's no need for you to post a reply to it at this point.

that's fine.

Reply 27

Zacken

Original post by RDKGames

To me it seems we have literally written the same thing. Maybe I just haven't explained it "by the book"

No, again:

\sqrt{4} \neq \pm 2

as you seem to think. That is plainly wrong.

Reply 28

1) Yes.

b^2=a \rightarrow b=\pm\sqrt{a}

(where a is whatever you want) but I'm unsure what you mean by "get it onto the other side"? If you square root b, then you square root the other side too, you don't move anything across.

2) - We have:

\frac{1}{\frac{R_1+R_2}{R_1R_2}}

- Multiply top and bottom by

R_1R_2

which gives:

\frac{1\cdot R_1R_2}{\frac{R_1+R_2}{R_1R_2} \cdot R_1R_2}

- Things cancel on the denominator and we are left with:

\frac{R_1R_2}{R_1+R_2}

3) Technically, you can just leave your answer as you found it firstly, but the answer in your book multiplied the fraction's numerator and denominator by -1 because it get rids of the negative on then numerator thus making it look simpler in a way.

You found it to be:

\frac{-fv}{f-v}

Multiplying top and bottom by -1 doesn't change the value because:

\frac{-fv}{f-v} \cdot 1 = \frac{-fv}{f-v} \cdot \frac{-1}{-1} = \frac{(-1)(-fv)}{(-1)(f-v)} = \frac{fv}{v-f}

The tricks for 2) and 3) are literally the same, I hope you can see how this is helpful in achieving the answers in your book. If you multiply the numerator and denominator by the same factor, the quantity is unchanged because that very factor would cancel out anyway, however rather than cancelling it out, you can multiply across with other terms on the numerator and denominator.

(edited 7 years ago)

Reply 29

I'm not a maths teacher, but I've done some maths tutoring before :smile:

1) If you square root a quantity, such as b there, then the other side requires to be the plus or minus of whatever you square rooted as well. As for the side on which b is, it doesn't require this sign.

2) It's the same idea as in 3) you just use

R_1R_2

rather than

-1

\rightarrow \frac{1}{\frac{R_1+R_2}{R_1R_2}} \cdot 1

=\frac{1}{\frac{R_1+R_2}{R_1R_2}} \cdot \frac{R_1R_2}{R_1R_2}

= \frac{1\cdot R_1R_2}{\frac{R_1+R_2}{R_1R_2} \cdot R_1R_2}

Notice within the main denominator the bottom of the smaller fraction can now be cancelled out because

\frac{a}{b} \cdot b = a

= \frac{R_1R_2}{R_1+R_2}

(edited 7 years ago)

Reply 30

miaofcourse

Original post by BasharAssad

since it's in the form of ax^2 + bx and not ax^2 +bx + c (i.e. no c term) simply take out the common factor 'x', so you are correct.

Thank you!

Reply 31

Degree? Hah, mate I only just finished Y13 this summer, going to study Maths for a degree in September though! I'm just used to teaching maths the way my teacher taught it, really. And unfortunately I am quite a distance from Brimingham but you can always ask here on the forums and people will explain stuff to you one way or another! :smile:

Reply 32

I've applied to Loughborough (firm) and Leicester so I'm waiting on results day to see what happens with those. :smile:

Reply 33

sindyscape62

It's your cancelling out that's causing the problem.

\frac{(3x+2)^2}{6x}

can't be simplified any further as there's no common factor- the same with he other fraction.
To simplify a fraction you must divide the top and bottom by the same thing. In your working with the first fraction you've divided the top by

3x+2

and the bottom by

x

These are not the same so you've changed the fraction.
With the second one you've divided the top by

x

and I'm not sure what you've done to the bottom. With this question multiplying together should be your first step.

Reply 34

sindyscape62

Ah, ok, sorry. That makes much more sense! What you've done is made a mistake with taking out the

3x+2

factor.
When you divide

6x+4

3x+2

you are left with just 2, not

2x+2

So after cancelling out you have

\frac{3x+2}{6} \times \frac{x^3}{2}

Hope that helps!

Reply 35

Yes, you divide the whole expressions as in general, (a+b)/(c+d) does not equal (a/c)+(b/d).

Reply 36