yes, but by completing the square we have divided by the 3(Original post by Bhaal85)
dy/dx = 6x+8 right.
set to equal 0 = 6x+8=0
rearrange to make x subject you get: x=8/6
sub that into orignal equation 3x^2+8x+16 = 10r2r3. (ten and two thirds)
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 41
 27102003 22:29

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 42
 27102003 22:30
(Original post by clever_lad)
oops, sorry Bhall, i didnt complete the square correctly. i am really sorry mate, i take my comment back *shrieks with embarrasment*........if you dont mind mate, please could you go through the process of completing the square for this equation, then showing me what x value will give the minimum value for the parabola.....please mate...once again, sorry. *shudders* 
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 43
 27102003 22:30
oops, sorry Bhall, i didnt complete the square correctly. i am really sorry mate, i take my comment back *shrieks with embarrasment*........if you dont mind mate, please could you go through the process of completing the square for this equation, then showing me what x value will give the minimum value for the parabola.....please mate..once again, sorry. *shudders*

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 44
 27102003 22:32
(Original post by theone)
yes, but by completing the square we have divided by the 3
We are using the original equation. 
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 27102003 22:34
(Original post by Bhaal85)
It makes no difference in this case, as we are not completing the square. Trust me mate. You diffeerentiate to find a min/max where the gradient = 0. Then sub the X coordinate into the orignal value to get the y value. Trust me on this.
We are using the original equation.
I'm claiming tha written in completed square form:
f(x) = 3((x+4/3)^2+32/9) from which we can deduce that ymin is when x = 4/3 and ymin = 3 .(32/9) = 32/3.
Glad we got that sorted out. 
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 46
 27102003 22:34
(Original post by clever_lad)
oops, sorry Bhall, i didnt complete the square correctly. i am really sorry mate, i take my comment back *shrieks with embarrasment*........if you dont mind mate, please could you go through the process of completing the square for this equation, then showing me what x value will give the minimum value for the parabola.....please mate..once again, sorry. *shudders*
3x^2 + 8x +16 = 3(x^2 + 8x/3 + 16/3) = 3((x+4/3)^2 + (16/316/9)) = 3((x+4/3)^2+32/9).
See above post for deductions of ymin and the xvalue. 
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 47
 27102003 22:36
aahhh i understand... its the "Scheitelpunktsformel"
might be right, but f'(x) is much more comfortable 
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 27102003 22:38
(Original post by Bhaal85)
Which method? Is it the basic one where the coefficient of x^2 is just one, or where its greater than one? 
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 49
 27102003 22:41
(Original post by ZJuwelH)
Both by any chance please??? Thanks! 
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 50
 27102003 22:48
(Original post by theone)
Well, if the coefficent of x is greater than 1, just divide through the whole quadratic by it and complete the square on your new quadratic using the method I wrote on the previous page about quadratics in the form of x^2+bx+c. 
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 51
 27102003 22:49
(Original post by ZJuwelH)
So e.g. 3x^2+6x12 becomes x^2+2x4, and complete the square of this? 
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 52
 27102003 22:50
(Original post by theone)
Oh, I'll go through it:
3x^2 + 8x +16 = 3(x^2 + 8x/3 + 16/3) = 3((x+4/3)^2 + (16/316/9)) = 3((x+4/3)^2+32/9).
See above post for deductions of ymin and the xvalue.
Basically for to put it into completed square form you must get it in the form a((xb)^2+q)
The original equation is 3x^2 + 8x +16.
Multiply the bracket out for a((xb)^2+q) this gives
ax^22apx+ap^2+aq compare to the original equation
ax^22apx+ap^2+aq : 3x^2 + 8x +16.
we can find that ax^2 is the same as 3x^2 from deduction we see that a=3
now we look at 2apx and 8x
2apx : 8x
we know that a = 3 therefore
2(3)px : 8x
6px :8x cancel the x's to get 6p:8 therfore p=8/6 = 4/3 (cancelled down)
now we look at the final bits:
16 and ap^2+aq, we know a=3 and p=4/3
therefore:
16 : (3)(4/3^2)+(3)q =
16 : 5r2r3 + 3q
16  5r2r3 = 3q therefore q = (10r2r3)/3 = 32/9
we know what a, p and q are therefore
a=3 and p=4/3 and q=32/9
sub all the bits into a((xp)^2+q) gives
3((x+4/3)^2+32/9)
Phew............................ ...........If anybody wants to give me rep for explaining all this, please feel free to do so.......... 
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 53
 27102003 22:51
One word of warning, you can only use this method to find the x value for the minimum.
When you have this, you have to put this back into f(x) to find the yvalue, you can not deduce it immediately from your completed square, since you have divided at the start. 
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 54
 27102003 22:53
(Original post by Bhaal85)
I concur. I actually wrote it out and worked it out.
Basically for to put it into completed square form you must get it in the form a((xb)^2+q)
The original equation is 3x^2 + 8x +16.
Multiply the bracket out for a((xb)^2+q) this gives
ax^22apx+ap^2+aq compare to the original equation
ax^22apx+ap^2+aq : 3x^2 + 8x +16.
we can find that ax^2 is the same as 3x^2 from deduction we see that a=3
now we look at 2apx and 8x
2apx : 8x
we know that a = 3 therefore
2(3)px : 8x
6px :8x cancel the x's to get 6p:8 therfore p=8/6 = 4/3 (cancelled down)
now we look at the final bits:
16 and ap^2+aq, we know a=3 and p=4/3
therefore:
16 : (3)(4/3^2)+(3)q =
16 : 5r2r3 + 3q
16  5r2r3 = 3q therefore q = (10r2r3)/3 = 32/9
we know what a, p and q are therefore
a=3 and p=4/3 and q=32/9
sub all the bits into a((xp)^2+q) gives
3((x+4/3)^2+32/9)
Phew............................ ...........If anybody wants to give me rep for explaining all this, please feel free to do so.......... 
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 55
 27102003 22:55
(Original post by Bhaal85)
I concur. I actually wrote it out and worked it out.
Basically for to put it into completed square form you must get it in the form a((xb)^2+q)
The original equation is 3x^2 + 8x +16.
Multiply the bracket out for a((xb)^2+q) this gives
ax^22apx+ap^2+aq compare to the original equation
ax^22apx+ap^2+aq : 3x^2 + 8x +16.
we can find that ax^2 is the same as 3x^2 from deduction we see that a=3
now we look at 2apx and 8x
2apx : 8x
we know that a = 3 therefore
2(3)px : 8x
6px :8x cancel the x's to get 6p:8 therfore p=8/6 = 4/3 (cancelled down)
now we look at the final bits:
16 and ap^2+aq, we know a=3 and p=4/3
therefore:
16 : (3)(4/3^2)+(3)q =
16 : 5r2r3 + 3q
16  5r2r3 = 3q therefore q = (10r2r3)/3 = 32/9
we know what a, p and q are therefore
a=3 and p=4/3 and q=32/9
sub all the bits into a((xp)^2+q) gives
3((x+4/3)^2+32/9)
Phew............................ ...........If anybody wants to give me rep for explaining all this, please feel free to do so.......... 
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 56
 27102003 22:55
(Original post by theone)
One word of warning, you can only use this method to find the x value for the minimum.
When you have this, you have to put this back into f(x) to find the yvalue, you can not deduce it immediately from your completed square, since you have divided at the start.
f(x)=3x^2+6x12 :3
1/3*f(x)=x^2+2x4
if you put the x value in the second you have 1/3 f(x)! 
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 57
 27102003 23:04
(Original post by Bhaal85)
Multiply the bracket out for a((xb)^2+q) this gives
ax^22apx+ap^2+aq compare to the original equation

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 58
 27102003 23:05
(Original post by Bhaal85)
This is the method, where you cant mess up, takes longer, but you do get it right. I personally use the other way, but I am not about to explain it cause my fingers are sore from typing all this. So if anybody wants to show their appreciation gimme some rep, cause I need it. Thanks.
I'll quickly go through the other method.
ax^2 + bx +c = a(x^2+bx/a+c/a).
Now we want to complete the square on x^2+bx/a+c/a.
Note that if x^2+bx/a+c/a = (x+m)^2 + n (where m and n are unknown). Since the only x term on the rhs is 2mx, we get b/a = 2m, and so m = b/2a (i.e. we jsut divide the coefficent of x by 2).
Now multiplying out, the constant terms on each side must be equal so m^2 + n = c/a. using m = b/2a we get n = c/a  b^2/4a^2.
And our completed square form is a((x+b/2a)^2+(c/a  b^2/4a^2)).
This is a lot harder to grasp with algebra than by just practice and an appreciation of the method.
Look away now if you don't want to see more

Using ax^2 + bx + c = a((x+b/2a)^2+(c/a  b^2/4a^2)).
If ax^2 + bx + c = 0, then a((x+b/2a)^2+(c/a  b^2/4a^2)) = 0
Now a is not 0 so we can say (x+b/2a)^2+(c/a  b^2/4a^2) = 0
So (x+b/2a)^2 = (b^24ac)/4a^2.
So x+b/2a = +/ (b^24ac)^(1/2)/2a and so x = (b +/ (b^24ac)^(1/2))/2a
Phew  Just thought I'd add that on since I'd done most the hard work anyway....
Time to return the favour Bhaal 
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 59
 27102003 23:08
(Original post by ZJuwelH)
Woah pullup! Where did p come from and where did b go? Did you mean p when you said b? 
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 60
 27102003 23:09
(Original post by theone)
I gave you some rep for just being bothered to type that much!.
I'll quickly go through the other method.
ax^2 + bx +c = a(x^2+bx/a+c/a).
Now we want to complete the square on x^2+bx/a+c/a.
Note that if x^2+bx/a+c/a = (x+m)^2 + n (where m and n are unknown). Since the only x term on the rhs is 2mx, we get b/a = 2m, and so m = b/2a (i.e. we jsut divide the coefficent of x by 2).
Now multiplying out, the constant terms on each side must be equal so n = m^2 + n = c/a. using m = b/2a we get n = c/a  b^2/4a^2.
And our completed square form is a((x+b/2a)^2+(c/a  b^2/4a^2)).
This is a lot harder to grasp with algebra than by just practice and an appreciation of the method.
Look away now if you don't want to see more

Using ax^2 + bx + c = a((x+b/2a)^2+(c/a  b^2/4a^2)).
If ax^2 + bx + c = 0, then a((x+b/2a)^2+(c/a  b^2/4a^2)) = 0
Now a is not 0 so we can say (x+b/2a)^2+(c/a  b^2/4a^2) = 0
So (x+b/2a)^2 = (b^24ac)/4a^2.
So x+b/2a = +/ (b^24ac)^(1/2)/2a and so x = (b +/ (b^24ac)^(1/2))/2a
Phew  Just thought I'd add that on since I'd done most the hard work anyway....
Time to return the favour Bhaal
Updated: October 28, 2003
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