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I can't find the difference of two squares. I don't understand :(

i can do simple ones like :2x^2+6x
but struggling to do ones like
2x^2+5x+2

3x^2+10x−8

15x^2+42x−9

Would be great if someone could go through how you solve them.
Thanks x

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Reply 1

Princess Bubbles

LikeAStar

i can do simple ones like :2x^2+6x
but struggling to do ones like
2x^2+5x+2

3x^2+10x−8

15x^2+42x−9

Would be great if someone could go through how you solve them.
Thanks x

Those aren't differences between two squares.
A difference between two sqaures has an X^2 term and a number term which is a square number.

Reply 2

boromir9111

2x^2 + 5x +2 i think you just complete the square..... and the following 2 as well... can use the quadratic formula as well.

Reply 3

Pheylan

you mean you can't complete the square?

Reply 4

goldsilvy

Looks as though you have to factorise those expressions.

Reply 5

divinelord

complete the square, factorise or use the quadratic formula.

Reply 6

didgeridoo12uk

i never really learnt to complete the square, its a pretty useless skill. factorising and using the quadratic formula are far more usefull.

got it now.

LikeAStar

i can do simple ones like :2x^2+6x
but struggling to do ones like
2x^2+5x+2

3x^2+10x−8

15x^2+42x−9

Would be great if someone could go through how you solve them.
Thanks x

Those are not difference of two squares.

I'll take it you mean factorising? Or completing the square? I'll do both.

For factorising you need to come up with your own method. If you really have no iea how to factorise, PM me and I'll teach you my method that always works. Here are the answers anyway

Spoiler

For completing the square, here is the process:

- Factorise everything through by the number in front of the x^2 ( "a" )

2x^2+5x+2 = 2[x^2+\frac{5}{2}x+1]

- Divide the co-eff of the x-term (we'll call this co-eff "b" ) by two, then put the first two terms into brackets as (x + b/2)^2, then SUBTRACT ((b/2)^2). Don't forget to put the constant "c" down to the next line. This may sound complicated, but it's really not.

2[(x+\frac{5}{4})^2-(\frac{5}{4})^2+1]

- Simply what's inside the square brackets

2[(x+\frac{5}{4})^2-(\frac{25}{16})+1]=2[(x+\frac{5}{4})^2-\frac{9}{16}]

- Multiply out by what's in front of the square brackets

2(x+\frac{5}{4})^2-\frac{9}{8}

And that's your final answer!

Reply 9

steve2005

didgeridoo12uk

i never really learnt to complete the square, its a pretty useless skill. factorising and using the quadratic formula are far more usefull.

You are not serious, are you?

Reply 10

didgeridoo12uk

steve2005

You are not serious, are you?

i'm totally serious,

you can use it to solve a quadratic, and find the turning point.

but in nearly all cases its easier to factorise/use the formula, and differentiate to get exactly the same information. and you're less likely to make mistakes

Reply 11

milliondollarcorpse

didgeridoo12uk

i'm totally serious, what possible use is completing the square?

How about translations of quadratic graphs for one?

Reply 12

didgeridoo12uk

milliondollarcorpse

How about translations of quadratic graphs for one?

that can be done in other more general ways, that will work for practically all graphs not just quadratics.

its just got far too specific an application to be of any real use. nobody would ever use it at uni

Reply 13

steve2005

didgeridoo12uk

I thought you were a student at Cambridge, however, I now think you are not at university studying mathematics. IF you are then I think something is not right with the state of education in the UK

Reply 14

didgeridoo12uk

steve2005

I thought you were a student at Cambridge, however, I now think you are not at university studying mathematics. IF you are then I think something is not right with the state of education in the UK

studying engineering, not maths.

and you've still failed to show me the point of it.

it's like the people the memorise every single formula that could be of any use. rather than just learning the main ones, and how to derive the rest from first principles

Reply 15

steve2005

didgeridoo12uk

You were the one that said that completing the square was not necessary or useful and not worth learning. I point out that it is covered in GCSE.

Reply 16

didgeridoo12uk

steve2005

You were the one that said that completing the square was not necessary or useful and not worth learning. I point out that it is covered in GCSE.

i'd personally multiply out the equation of a circle

(x-a)^2 + (y-b)^2 = r^2

and compare coefficients.

might take slightly longer, but its one less thing i'd have to remember for an exam.

Reply 17

DFranklin

didgeridoo12uk

i'd personally multiply out the equation of a circle

(x-a)^2 + (y-b)^2 = r^2

and compare coefficients.

might take slightly longer, but its one less thing i'd have to remember for an exam.

I see no difference between doing that and completing the square. The important part of completing the square is knowing that you can rewrite x^2+bx+c in the form (x+B)^2+C for some B, C. Knowing the actual formulae for B and C is unimportant, but just because you don't bother to remember them doesn't mean you're not completing the square.

Reply 18

Liverpool F.C.™

Completing the square can also be used for finding the minimum/maximum points for a Quadratic. Not sure now if there is an alternative method.

Reply 19

milliondollarcorpse

didgeridoo12uk

i'd personally multiply out the equation of a circle

(x-a)^2 + (y-b)^2 = r^2

and compare coefficients.

might take slightly longer, but its one less thing i'd have to remember for an exam.

All you need to 'remember' is that you're putting the quadratic into the form of a perfect square (with a constant outside). There's no formula or anything.

Liverpool F.C.&#8482

Completing the square can also be used for finding the minimum/maximum points for a Quadratic. Not sure now if there is an alternative method.