Leaving cert maths

Maria656

What grade should I be getting in 5th year in order to stay in honours maths ??

LCstudenthelp

I need someone to explain this to me I don’t get it at all!!

2x^2 - 7x - 10
2(x^2 -7/2x) - 10
2(x -7/4)^2 - 49/8 - 10
2(x - 7/4)^2 - 129/8

Thank you

ignorance is strength

LCstudenthelp wrote: »

I need someone to explain this to me I don’t get it at all!!

2x^2 - 7x - 10
2(x^2 -7x/2) - 10
2(x -7/4)^2 - 49/8 - 10
2(x - 7/4)^2 - 129/8

Thank you

What do you need explained? How you get from the first line to the fourth?


2x^2 - 7x - 10


[2x^2 - 7x] - 10 ...factorise the brackets by dividing both elements by 2...

1 2[x^2 - 7x/2] - 10

A^2 + 2AB + B^2 = (A + (A +

=> A^2 + 2AB = (A + (A + - B^2

A^2 + 2AB <=> x^2 - 7x/2

(A = x, 2AB = -7x/2, B=...)

2 [x^2 - 7/2x] = (A + (A + - B^2 = (x - 7/4)(x - 7/4) - 49/16

2[(x - 7/4)(x - 7/4) - 49/16] - 10 ...subbing 2 into 1...

2(x - 7/4)^2 - 129/8



That's a bit of a mess and makes it seem more complicated than it is. Hope it helps.

LCstudenthelp

ignorance is strength wrote: »

What do you need explained? How you get from the first line to the fourth?


2x^2 - 7x - 10


[2x^2 - 7x] - 10 ...factorise the brackets by dividing both elements by 2...

1 2[x^2 - 7x/2] - 10

A^2 + 2AB + B^2 = (A + (A +

=> A^2 + 2AB = (A + (A + - B^2

A^2 + 2AB <=> x^2 - 7x/2

(A = x, 2AB = -7x/2, B=...)

2 [x^2 - 7/2x] = (A + (A + - B^2 = (x - 7/4)(x - 7/4) - 49/16

2[(x - 7/4)(x - 7/4) - 49/16] - 10 ...subbing 2 into 1...

2(x - 7/4)^2 - 129/8



That's a bit of a mess and makes it seem more complicated than it is. Hope it helps.

Thank you that does help but I still don’t understand where the x goes from 7/2x

ignorance is strength

LCstudenthelp wrote: »

Thank you that does help but I still don’t understand where the x goes from 7/2x

2AB = -7x/2

...so B = -7x/2 / 2A

...and because A = x, B = -7/4.


I don't quite know what you mean by "where x goes" and I'm not sure I can explain it either. x is detached from the coefficient when you factorise it. They're still effectively multiplied together, it's just that in order to see that you need to remove the brackets.

derb12

There are a couple of underlying tricks in this that you need to understand before you can attempt a question like the above.
For example can you do this

(x-3)^2

A lot of students get this wrong, so I'm just making sure you have the basics. Some remember this as "square the first, square the second, twice their product".

The answer is x^2 -6x + 9

Knowing where the -6x comes from is key to understanding where the (x-7/4)^2 bit comes from above.

So now if you are starting with something like

x^2 + 10x - 3

you just look at the first two terms (park the -3) and try to write x^2 + 10x in the form (x ....)^2

If you know the rule above you'll see that your answer will start with (x+5)^2

But when you expand (x+5)^2 you get x^2 + 10x + 25

So that means that we will end up with the following, starting from the top

x^2 + 10x -3
= (x +5)^2 -25 -3 (The -25 is there to make up for the +25 we introduced by squaring (x+5)
=(x+5)^2 -28


Your example above is complicated by the coefficient of x^2
It might be easier to follow if you insert the line

2((x -7/4)^2 - 49/16) - 10

before
2(x -7/4)^2 - 49/8 - 10

Look up "completed square form" - if you are using Active Maths4 it's on page 170