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1. algebra problem
it's part of another question (on coordinate geometry) but the part i'm stuck on is to find the point the crosses the parabola y^2 = 16x and the straight line 4x-3y -16 = 0

they have already given us the point (16,16) so we need to find the other one where they meet:

my working:

y^2 = 16x ==> y = 4x^(1/2)
4x -3(4x^(1/2)) - 16 = 0
4x - 12x^(1/2) - 16 = 0
x - 3x^(1/2) - 4 = 0
x^2 - 9x - 16 = 0
x(x-9) = 16

so x = 16 which is given to us, however the correct answer for the other x value is 1, not 25

could anyone show me where i'm going wrong?
2. Re: algebra problem
(Original post by lemonpwns1)
x - 3x^(1/2) - 4 = 0
x^2 - 9x - 16 = 0
The second line is a mistake. You can't simplify an equation by squaring each term.

If you follow the algebra rule "Do the same to the LHS as the RHS" then you can square both sides to get:

But your mistake was in saying that e.g. , which is not true in general.

It's never a good idea to involve a square root in an equation so that's what's lead to your problems. A better method is to rearrange the second equation so that it is in terms of x (or 16x) and substitute that into the first equation.

Try this and then post your working if you're still stuck.

Edit You made a second mistake after your first error. You said that x(x-9)=16 implies that x=16 but this is false. You can only use this method if the right-hand-side is 0 e.g. x(x-9)=0 implies that x=0 or x=9 but e.g. x(x-9)=23 does not imply that x=23 and x=32.

To solve the equation , you would need to use the quadratic formula since it cannot be factorised. But since the equation is false (from the previous error), you don't need to try this.
Last edited by notnek; 23-07-2012 at 05:39.
3. Re: algebra problem
(Original post by notnek)
The second line is a mistake. You can't simplify an equation by squaring each term.

If you follow the algebra rule "Do the same to the LHS as the RHS" then you can square both sides to get:

But your mistake was in saying that e.g. , which is not true in general.

It's never a good idea to involve a square root in an equation so that's what's lead to your problems. A better method is to rearrange the second equation so that it is in terms of x (or 16x) and substitute that into the first equation.

Try this and then post your working if you're still stuck.

Edit You made a second mistake after your first error. You said that x(x-9)=16 implies that x=16 but this is false. You can only use this method if the right-hand-side is 0 e.g. x(x-9)=0 implies that x=0 or x=9 but e.g. x(x-9)=23 does not imply that x=23 and x=32.

To solve the equation , you would need to use the quadratic formula since it cannot be factorised. But since the equation is false (from the previous error), you don't need to try this.
I already done it by getting it in terms of x thats why I posted, but thank you I managed to get it by doing

x-3x^(1/2) -4 = 0
x^(1/2) = [(4-x)/(-3)]^2

and then solving for x = 16 and x = 1

cheers