Maths Core 1- question

lynchy6789 · 14 May 2009 at 21:30

Got core 1 exam coming up soon and go through past papers and everything i find pretty reasonable but there is one thing that i can't get my head around and that "Equations which reduce to quadratic equations".

This is what i am stuck on OCR June 07 paper

" that mean to the power of i.e. "2 means squared.

By using substitution y=(X+2)"2 find the real roots of the equation

(X+2)"4+5(X+2)-6=0

This is what i did (Wrong)

(Y+2)"2+5(Y+2)-6

Y"2+4Y+4+5Y+10-9=0

Y2+9Y+8

(Y+8)(Y+1)

But this is wrong as the answer
y2 + 5y – 6 = 0
(y + 6)(y - 1) = 0

Past Paper question

Thanks
Answer paper

lynchy6789 · 14 May 2009 at 21:39

sorry yes its (X+2)^4+5(X+2)^2-6=0

How do you get Y^2+5Y-6

I know how to solve it just i dont see how you get the above as when you put it =Y

I get (Y+2)^2+5(Y+2)-6

that gives Y^2+4y+4+5y+10-6 <<<<<< Here is where i am going wrong but dont know why?

lynchy6789 · 14 May 2009 at 21:48

Vixen said:
Using the terms from your example, substitution means whenever you see (x+2)^2 in the given equation you write y instead. So (x+2)^4 = ((x+2)^2)^2 = y^2, etc

Thanks this really helped me i just presumed that as its (x+2)^4 you replace X with Y and and make it ^2.

Probably why i always struggled with them.

so if we had X^2/3+3X1/3-10=0

would Y= X^1/3

((X^1/3)^1/3+((3x)^1/3-10=0

Y^2+3Y-10?

lynchy6789 · 14 May 2009 at 21:57

not only a computer forum also a maths forum

.

Actually i have now confusled myself. Y= X^1/3

((X^1/3)^1/3) >>>>>>>>>>>>>> ((Y)^1/3)>>>>> wont that make it Y^1/3 why is Y^2 :S

lynchy6789 · 14 May 2009 at 22:11

Vixen said:
Look back at what you wrote initially, it's slightly different.

X^2/3 is the same as (x^1/3)^2.

EDIT: you put a bracket in the wrong place on the expansion, it should read (x^1/3)^2 + 3(x^1/3) - 10 = 0.

I think you're trying to make it more complicated than it actually is. Go through each equation looking for the term you're substituting and put it in brackets. Then use rules of powers to figure out what's missing. In the first example you had (x+2)^4 and y = (x+2)^2, so how do you make 4 from 2? (x+2)^2x2 is the same as ((x+2)^2)^2.
In the second one, you've got x^2/3 and y = x^1/3, how do you make 1/3 into 2/3? It's x^((1/3)x2).

If you ever struggle to remember how powers work, go back to powers of 2. So in the first example, how do you make 2^(2x2) = 2^4? You've got real numbers to plug into a calculator, and just apply the same principle.

That has really cleared things up i think i need to stop missing out steps as that where i am making mistakes. I have Core 1 this wednesday and Core 2 friday

lynchy6789 · 14 May 2009 at 22:52

as level

lynchy6789 · 15 May 2009 at 00:04

so far i am findng core 2 a little easier i have always done better in calculator papers.

Need to make sure i don't rush and make silly mistakes.

lynchy6789 · 15 May 2009 at 00:29

i am going to do core 1 core 2 core 3 core 4 decision 1 and Mechanics 1

lynchy6789 · 15 May 2009 at 00:49

Yea also luckily it is really late June 17th so means I will have a good 2 weeks solid revising that. Few tricky areas like simplex! Next year at a2 goi g to do a physics as instead at our 6th form general studies