Maths Core 1- question

lynchy6789

lynchy6789

lynchy6789

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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
 
By using substitution y=(X+2)^2

(X+2)^4+5(X+2)-6=0 <-- are you sure?

(X+2)^4+5(X+2)^2-6=0 <-- maybe this?

then:
y^2 + 5y - 6 = 0. thats easy to solve, yes?

edit: more verbose:
Code: 
(X+2)^4+5(X+2)^2-6 = [(x+2)^2]^2 + 5[(x+2)^2] - 6
what is (x+2)^2 ? y!
therefore
Code: 
[(x+2)^2]^2 + 5[(x+2)^2] - 6 = y^2 + 5y - 6
[    y  ]^2 + 5[   y   ] - 6  = 0
 
Last edited:
when you let y = (x+2)^2 your equation then becomes y^2 + 5y - 6

Edit: Beaten

Edit 2: Damn you Amleto, was going to give a long explaination and you beat me again :p
 
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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
 
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?
 
you are substituting wrong. you arent doing what it tells you.

you can check yours is wrong - if you replace all y in your equation with (x+2)^2 you should get back the original form. but you get something massively different.
 
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Y is equal to all of (x+2)^2 therefore everytime you see that replace the whole thing with Y.

Think of Y as being a box which you can fix that term and only that term into, and then, in the name of tidyness, everytime you see that term you put it into it's box. The contents are still there but they're now in a nice friendly Y shaped box.
 
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Yeah you replace the WHOLE of (x+2)^2 with y, not just the x term within it.

So it becomes y rather than (y+2)^2.
 
lynchy6789 said:
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?
Click to expand...

Well since Y=(X+2)^2, then (X+2)^4 must be Y^2.

5(X+2)^2 is the same as 5Y. You just swap the (X+2)^2 out for Y.

Then you have the left over -6.

So put it all together, and you get Y^2 + 5Y - 6
 
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
Click to expand...

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 said:
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?
Click to expand...

Yes, that's right. Glad you understand it a bit better :).
 
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 said:
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
Click to expand...

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.
 
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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.
Click to expand...


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 :o
 
lynchy6789 said:
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 :o
Click to expand...

It always sounds like a good idea to skip steps as it saves time, but I've always found that working through every step means you make fewer mistakes :).
 
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