Need Urgent Help Simplifying Quadradics (1 Viewer)

[ASDFGHJKL:""]

I am having problems simplifying quadradics e.g:

3n[squared]+5n-1062 = 0

(3n+59)(n-18)=0

I got the example out of the text book. I used to be alright with smaller and simpler examples but now that they are more complicated I have no idea of what to do. Can someone please explain how you/what methods you use to simplify these expressions.

Thanks in advance


<!--[if !msEquation]-->[FONT=&quot]<!--[if gte vml 1]><v:shapetype id="_x0000_t75" coordsize="21600,21600" o:spt="75" oreferrelative="t" path="m@4@5l@4@11@9@11@9@5xe" filled="f" stroked="f"> <v:stroke joinstyle="miter"/> <v:formulas> <v:f eqn="if lineDrawn pixelLineWidth 0"/> <v:f eqn="sum @0 1 0"/> <v:f eqn="sum 0 0 @1"/> <v:f eqn="prod @2 1 2"/> <v:f eqn="prod @3 21600 pixelWidth"/> <v:f eqn="prod @3 21600 pixelHeight"/> <v:f eqn="sum @0 0 1"/> <v:f eqn="prod @6 1 2"/> <v:f eqn="prod @7 21600 pixelWidth"/> <v:f eqn="sum @8 21600 0"/> <v:f eqn="prod @7 21600 pixelHeight"/> <v:f eqn="sum @10 21600 0"/> </v:formulas> <vath o:extrusionok="f" gradientshapeok="t" o:connecttype="rect"/> <o:lock v:ext="edit" aspectratio="t"/> </v:shapetype><v:shape id="_x0000_i1025" type="#_x0000_t75" style='width:6pt; height:15pt'> <v:imagedata src="file:///C:\Users\Andrew\AppData\Local\Temp\msohtmlclip1\01\clip_image001.png" o:title="" chromakey="white"/> </v:shape><![endif]--><!--[if !vml]--><!--[endif]-->[/FONT]<!--[endif]-->

 

[gh0stface]

use the quadratic formula: [-b (+or-) squareroot (b^2-4ac)]/2a

 

[m.incognito]

ya.

 

[ASDFGHJKL:""]

thanks for your help

 

[PC]

That's a nasty question.

You would have been shown at least one way to factorise quadratic expressions, most commonly is the "cross" method.

Solve 3n² + 5n - 1062 = 0.

So we need two numbers that multiply to give the first term 3n².
We also need two numbers that multiply to give the last term -1062.

The first bit will be easy, they have to be 3n and n. 3n x n = 3n². Put these two numbers in the cross.

3n\/
n/\

The second bit is harder. Factors of 1062 are 1, 2, 3, 6, 9, 18, 59, 118, 177, 354, 531 , 1062. We need to put two of these numbers in the cross. BUT we also need to include a positive and a negative somewhere because the number is actually -1062.

For example,

3n\/177
n/\–6

Now we multiply across the crosses and add.
3n x –6 = –18n and n x 177 = 177n, so –18n + 177n = 159n
But we need a middle term of +5n.
This combination of 177 and –6 doesn't work.

A good strategy is that because we need +5n, the bigger number needs to be positive.

Try this one:
3n\/59
n/\–18

So across the cross:
3n x -18 = –54n and n x 59 = 59n, and –54n + 59n = 5n.
That's what we need.

So the factors are (3n + 59) and (n - 18).

To solve, put each factor to zero.

3n + 59 = 0
3n = –59
n = -19 2/3

and

n - 18 = 0
n = 18

So the solutions are n = -19 2/3 and 18

Hope this helps

 

[Caitlin63]

ok, these are nasty numbers so if I saw this question in a test i would use the quadratic formula as it would not take much longer.

However to factorise... yes as above post states the cross method is the most common way students are taught these days but I was taught using a different method which is very effective for all co-efficients of x. I thought I would post my working to this question for anyone who struggles with the cross method or is looking for an alternative.

3n^2 + 5n - 1062 = 0

(3n ....) (3n... ) = 0​

____________
3
(this is the same as for simple quadratics but to account for the two lots of 3n we will divide by 3)

Note: when using this method you must also multiply the constant (number with no "n"s in it) by the number you are dividing by). In this question -1062 becomes -3186.

Next step is to find two numbers that multiply to give -3186 and add to give 5. This means the larger factor needs to be positive as a positive number remains when the two are added. So the next step of working is...

(3n + 59) (3n - 54) = 0 (factors found are 59 and - 54)
_______________
3

Next step is to take out a common factor in one (or both - depends on question) set(s) of brackets on top of the fraction

(3n + 59) 3(n - 18) = 0
_______________
3

The three on the top will now cancel with the 3 on the bottom leaving

(3n + 59) (n - 18) = 0

The two solutions are given by

3n + 59 = 0
3n = -59
n = - 19 2/3

AND
n - 18 = 0
n = 18
Hope someone out there finds this interesting or useful.

 

[Caitlin63]

Yup the formula will be much faster to use, but yes if you had lots of time you can factorise to check