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# C1 maths coordinate question? watch

1. I am stuck on part c of this question. Can you please help me?? A step by step explanation would be useful!

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2. if you have

x(a quadratic in x) = 0

then either x = 0 or the quadratic = 0

since they want surds in the answer it is not likely that the quadratic will factorise.
3. (Original post by the bear)
if you have

x(a quadratic in x) = 0

then either x = 0 or the quadratic = 0

since they want surds in the answer it is not likely that the quadratic will factorise.
how am i supposed to find the coordinates?
4. You need to start by finding the roots of the equation x(x^2-8x+4). This can be found by using the quadratic formula on the inside part of the bracket. This will then give you three answers: 0 (because of the x outside of the brackets) and two surds. The positive one is the x-coordinate.

Next, you need to put the x-coordinate into one of the original formulae in order to work out the y-coordinate.

Hope this helps!
5. (Original post by ayylmao97)
how am i supposed to find the coordinates?
you get the x coordinates by solving the equation, then to find the y coordinates you put the x coordinates into one of the two original functions ( it does not matter which ).
6. Did you get the answer?
7. (Original post by Student 977)
no, i don't know how to solve the equation
8. (Original post by ayylmao97)
no, i don't know how to solve the equation
You need to use the quadratic formula on (x^2-8x+4). This will give you two answers as surds. Can you do that?
9. (Original post by Student 977)
You need to use the quadratic formula on (x^2-8x+4). This will give you two answers as surds. Can you do that?
so far i got to (8+-√48)/2 but i don't know how to cancel down the root
10. (Original post by ayylmao97)
so far i got to (8+-√48)/2 but i don't know how to cancel down the root
I forgot it was a non-calculator paper! You need to complete the square of x^2-8x+4.

11. Here is some help!
12. (Original post by Student 977)
I forgot it was a non-calculator paper! You need to complete the square of x^2-8x+4.
I've done it, and I got (x-4)^2-12
13. (Original post by ayylmao97)
I've done it, and I got (x-4)^2-12
Yes and then you can equate that to x
14. (Original post by Student 977)
Yes and then you can equate that to x
After that do I substitute the x coordinate into one of the equations to find y?
15. (Original post by ayylmao97)
After that do I substitute the x coordinate into one of the equations to find y?
Yes, but make sure you use the correct x-coordinate.
16. (Original post by Student 977)
Yes, but make sure you use the correct x-coordinate.
Lool sorry to keep bothering you like this but I don't know how to expand the brackets once I've substituted x
17. Which equation did you put the x into?
18. (Original post by Student 977)
Which equation did you put the x into?
Into the x(4-x)
19. So you get (4-2rt3)(4-4-2rt3)
20. (Original post by Student 977)
So you get (4-2rt3)(4-4-2rt3)
yes but then wouldnt the second bracket cancel down to (4-2rt3)?

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