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Algebra/Factorising Question (or at least I think it is :s) - HSN forum

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Algebra/Factorising Question (or at least I think it is :s)


8 replies to this topic

#1 Adam

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Posted 15 March 2006 - 08:18 PM

Hey, can somebody complete & explain how this question is done? I'd be very greatful, cheers:

IPB Image
(Area of the first rectangle is A1 and area of the second is A2)

A1 - A2 = x2 - (8p + 4)x - 8p

Given that A1-A2 = 1cm2, establish the value of p, where p>-1, for this equation to have only one solution for x.

And hence find x when p takes this value.

#2 dfx

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Posted 15 March 2006 - 10:15 PM

 A_1 - A_2 = x^2 - (8p + 4)x - 8p = 1cm^2 
\\
x^2 - 8px -4x - 8p - 1 = 0 
\\
x^2 - (8p + 4)x - (8p + 1) = 0

This equation will only have one solution (of x) if the discriminant = 0 i.e. real and equal roots (effectively one root).

So solve for  b^2 - 4 a c = 0 and get your value of p.

And then just plug it back in for the value of x and solve the original quadratic.

#3 Adam

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Posted 16 March 2006 - 06:46 PM

QUOTE(dfx @ Mar 15 2006, 10:15 PM) View Post

 A_1 - A_2 = x^2 - (8p + 4)x - 8p = 1cm^2 
\\
x^2 - 8px -4x - 8p - 1 = 0 
\\
x^2 - (8p + 4)x - (8p + 1) = 0

This equation will only have one solution (of x) if the discriminant = 0 i.e. real and equal roots (effectively one root).

So solve for  b^2 - 4 a c = 0 and get your value of p.

And then just plug it back in for the value of x and solve the original quadratic.

That rings a bell - cheers. I'm confused about this line though:

x^2 - (8p + 4)x - (8p + 1) = 0
How did you manage to arrive at this? smile.gif

Edit: Bleh, not sure if I'm doing this right...


b^2 - 4ac
\\
\\
a = 1 
\\b = -8 (8p + 4) 
\\c = - (8p + 1)
\\
\\
(-8 (8p + 4))^2 - (4(1)(-(8p + 1)))

Correct so far?

#4 Dave

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Posted 16 March 2006 - 07:31 PM

the line you dont understand comes about because there is a common factor of X in there. This makes sense as a quadratic is in the form Ax2+bx + c. noteing that p is just a number so -8p-1 is a constant

and b = -(8p+4)

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#5 Adam

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Posted 16 March 2006 - 08:06 PM

QUOTE(Dave @ Mar 16 2006, 07:31 PM) View Post

the line you dont understand comes about because there is a common factor of X in there. This makes sense as a quadratic is in the form Ax2+bx + c. noteing that p is just a number so -8p-1 is a constant

and b = -(8p+4)

Ah, yeah, silly me. I wrote it down correct when I was doing the calculation but typed it out wrong. tongue.gif

Anyhoo, I got: p = -20/8 and p = -4 for some reason.


a = 1
\\
b = - (8p + 4)
\\
c = - (8p + 1)
\\
\\
( - (8p + 4) )^2 - (4 (1) ( - (8p + 1)))
\\
(64p^2 + 64p + 16) - (- 32p - 4)
\\
64p^2 + 64p + 16 + 32p + 4
\\
64p^2 + 96p + 20
\\
64p^2 + 96p = -20
\\
8p (8p + 12) = -20
\\
\\
8p = -20
\\
p = -20/8
\\
\\
8p + 12 = -20
\\
8p = -32
\\
p = -4

Er, what did I do wrong this time?

#6 ad absurdum

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Posted 16 March 2006 - 09:18 PM

QUOTE(Betinho @ Mar 16 2006, 08:06 PM) View Post

QUOTE(Dave @ Mar 16 2006, 07:31 PM) View Post

the line you dont understand comes about because there is a common factor of X in there. This makes sense as a quadratic is in the form Ax2+bx + c. noteing that p is just a number so -8p-1 is a constant

and b = -(8p+4)

Ah, yeah, silly me. I wrote it down correct when I was doing the calculation but typed it out wrong. tongue.gif

Anyhoo, I got: p = -20/8 and p = -4 for some reason.


a = 1
\\
b = - (8p + 4)
\\
c = - (8p + 1)
\\
\\
( - (8p + 4) )^2 - (4 (1) ( - (8p + 1)))
\\
(64p^2 + 64p + 16) - (- 32p - 4)
\\
64p^2 + 64p + 16 + 32p + 4
\\
64p^2 + 96p + 20
\\
***64p^2 + 96p = -20
\\
8p (8p + 12) = -20
\\
\\
8p = -20
\\
p = -20/8
\\
\\
8p + 12 = -20
\\
8p = -32
\\
p = -4

Er, what did I do wrong this time?

I put *** where I think you went wrong - well this line is still correct but it the line following it is wrong. Keep the 64p^2 + 96p + 20 equal to zero, take out a common factor of four to get 4(16p^2 + 96p + 20) = 0. Now you have to split the part inside the bracket into two different brackets, and then find the values for p which make the brackets, and therefore the whole thing, equal to zero. One of them should be greater than -1.
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#7 Adam

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Posted 16 March 2006 - 11:51 PM

QUOTE(ad absurdum @ Mar 16 2006, 09:18 PM) View Post

QUOTE(Betinho @ Mar 16 2006, 08:06 PM) View Post

QUOTE(Dave @ Mar 16 2006, 07:31 PM) View Post

the line you dont understand comes about because there is a common factor of X in there. This makes sense as a quadratic is in the form Ax2+bx + c. noteing that p is just a number so -8p-1 is a constant

and b = -(8p+4)

Ah, yeah, silly me. I wrote it down correct when I was doing the calculation but typed it out wrong. tongue.gif

Anyhoo, I got: p = -20/8 and p = -4 for some reason.


a = 1
\\
b = - (8p + 4)
\\
c = - (8p + 1)
\\
\\
( - (8p + 4) )^2 - (4 (1) ( - (8p + 1)))
\\
(64p^2 + 64p + 16) - (- 32p - 4)
\\
64p^2 + 64p + 16 + 32p + 4
\\
64p^2 + 96p + 20
\\
***64p^2 + 96p = -20
\\
8p (8p + 12) = -20
\\
\\
8p = -20
\\
p = -20/8
\\
\\
8p + 12 = -20
\\
8p = -32
\\
p = -4

Er, what did I do wrong this time?

I put *** where I think you went wrong - well this line is still correct but it the line following it is wrong. Keep the 64p^2 + 96p + 20 equal to zero, take out a common factor of four to get 4(16p^2 + 96p + 20) = 0. Now you have to split the part inside the bracket into two different brackets, and then find the values for p which make the brackets, and therefore the whole thing, equal to zero. One of them should be greater than -1.


I think you're right - I tried that and got an answer of -0.25, so erm, yea. Thanks everybody. smile.gif

#8 John

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Posted 17 March 2006 - 01:14 PM

QUOTE(ad absurdum @ Mar 16 2006, 09:18 PM) View Post

QUOTE(Betinho @ Mar 16 2006, 08:06 PM) View Post

QUOTE(Dave @ Mar 16 2006, 07:31 PM) View Post

the line you dont understand comes about because there is a common factor of X in there. This makes sense as a quadratic is in the form Ax2+bx + c. noteing that p is just a number so -8p-1 is a constant

and b = -(8p+4)

Ah, yeah, silly me. I wrote it down correct when I was doing the calculation but typed it out wrong. tongue.gif

Anyhoo, I got: p = -20/8 and p = -4 for some reason.


a = 1
\\
b = - (8p + 4)
\\
c = - (8p + 1)
\\
\\
( - (8p + 4) )^2 - (4 (1) ( - (8p + 1)))
\\
(64p^2 + 64p + 16) - (- 32p - 4)
\\
64p^2 + 64p + 16 + 32p + 4
\\
64p^2 + 96p + 20
\\
***64p^2 + 96p = -20
\\
8p (8p + 12) = -20
\\
\\
8p = -20
\\
p = -20/8
\\
\\
8p + 12 = -20
\\
8p = -32
\\
p = -4

Er, what did I do wrong this time?

I put *** where I think you went wrong - well this line is still correct but it the line following it is wrong. Keep the 64p^2 + 96p + 20 equal to zero, take out a common factor of four to get 4(16p^2 + 96p + 20) = 0. Now you have to split the part inside the bracket into two different brackets, and then find the values for p which make the brackets, and therefore the whole thing, equal to zero. One of them should be greater than -1.


But if you take out a factor of 4 you get 4(16p^2 + 24p + 5)

#9 ad absurdum

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Posted 18 March 2006 - 10:39 AM

QUOTE(Sodium_Ho @ Mar 17 2006, 01:14 PM) View Post

QUOTE(ad absurdum @ Mar 16 2006, 09:18 PM) View Post

QUOTE(Betinho @ Mar 16 2006, 08:06 PM) View Post

QUOTE(Dave @ Mar 16 2006, 07:31 PM) View Post

the line you dont understand comes about because there is a common factor of X in there. This makes sense as a quadratic is in the form Ax2+bx + c. noteing that p is just a number so -8p-1 is a constant

and b = -(8p+4)

Ah, yeah, silly me. I wrote it down correct when I was doing the calculation but typed it out wrong. tongue.gif

Anyhoo, I got: p = -20/8 and p = -4 for some reason.


a = 1
\\
b = - (8p + 4)
\\
c = - (8p + 1)
\\
\\
( - (8p + 4) )^2 - (4 (1) ( - (8p + 1)))
\\
(64p^2 + 64p + 16) - (- 32p - 4)
\\
64p^2 + 64p + 16 + 32p + 4
\\
64p^2 + 96p + 20
\\
***64p^2 + 96p = -20
\\
8p (8p + 12) = -20
\\
\\
8p = -20
\\
p = -20/8
\\
\\
8p + 12 = -20
\\
8p = -32
\\
p = -4

Er, what did I do wrong this time?

I put *** where I think you went wrong - well this line is still correct but it the line following it is wrong. Keep the 64p^2 + 96p + 20 equal to zero, take out a common factor of four to get 4(16p^2 + 96p + 20) = 0. Now you have to split the part inside the bracket into two different brackets, and then find the values for p which make the brackets, and therefore the whole thing, equal to zero. One of them should be greater than -1.


But if you take out a factor of 4 you get 4(16p^2 + 24p + 5)
laugh.gif Oops, I can never put down what I mean when I'm typing maths. Thanks for pointing that out.

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