Sxv_
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How do I post the pic of the question
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Sxv_
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(Original post by Sxv_)
How do I post the pic of the question
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3317752
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you had  -2t^2 + 6t + 1 = 0

Divide the entire equation by -2 to get the coefficient of  t^2 to be 1

so u got  -2(t^2 - 3t - 0.5) = 0

then complete the square of whats inside the  ()

so you got  -2 [(t-1.5)^2 - 2.75] = 0

then multiple the entire expression by 2 to remove the second brackets

 -2(t-1.5)^2 -5.5 = 0

from this you can see that the points are  t = 1.5 h = 5.5
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Balkaran
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(Original post by Pretish)
you had  -2t^2 + 6t + 1 = 0

Divide the entire equation by -2 to get the coefficient of  t^2 to be 1

so u got  -2(t^2 - 3t - 0.5) = 0

then complete the square of whats inside the  ()

so you got  -2 [(t-1.5)^2 - 2.75] = 0

then multiple the entire expression by 2 to remove the second brackets

 -2(t-1.5)^2 -5.5 = 0

from this you can see that the points are  t = 1.5 h = 5.5
That would be wrong I think
 -2 [(t-1.5)^2 - 2.75] = 0
You have to multiply - 2.75 by - 2
Giving you this answer
 -2(t-1.5)^2 +5.5 = 0
Then you take the 5.5 over
 -2(t-1.5)^2 =  -5.5
Divide it by -2
 -(t-1.5)^2 = 2.75
Making the points
 t = 1.5 h = 2.75
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Integer123
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(Original post by Balkaran)
That would be wrong I think
 -2 [(t-1.5)^2 - 2.75] = 0
You have to multiply - 2.75 by - 2
Giving you this answer
 -2(t-1.5)^2 +5.5 = 0
Then you take the 5.5 over
 -2(t-1.5)^2 =  -5.5
Divide it by -2
 -(t-1.5)^2 = 2.75
Making the points
 t = 1.5 h = 2.75
(Original post by Pretish)
you had  -2t^2 + 6t + 1 = 0

Divide the entire equation by -2 to get the coefficient of  t^2 to be 1

so u got  -2(t^2 - 3t - 0.5) = 0

then complete the square of whats inside the  ()

so you got  -2 [(t-1.5)^2 - 2.75] = 0

then multiple the entire expression by 2 to remove the second brackets

 -2(t-1.5)^2 -5.5 = 0

from this you can see that the points are  t = 1.5 h = 5.5
When you find the minimum or maximum (vertex) of a quadratic, you don't set it equal to zero, as it isn't being solved. You complete the square, as was done correctly, giving

h = -2(t-1.5)^2 + 5.5

It may help to write this as h = 5.5 -2(t-1.5)^2

Now, the maximum value of h is achieved when the smallest value of 2(t-1.5)^2 is subtracted from 5.5. As the bracket is squared, the smallest value to be subtracted is zero. Hence the maximum of h is 5.5 and it is achieved when t = 1.5.
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Balkaran
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(Original post by Integer123)
When you find the minimum or maximum (vertex) of a quadratic, you don't set it equal to zero, as it isn't being solved. You complete the square, as was done correctly, giving

h = -2(t-1.5)^2 + 5.5

It may help to write this as h = 5.5 -2(t-1.5)^2

Now, the maximum value of h is achieved when the smallest value of 2(t-1.5)^2 is subtracted from 5.5. As the bracket is squared, the smallest value to be subtracted is zero. Hence the maximum of h is 5.5 and it is achieved when t = 1.5.
I didnt really know how to do it. I just went off the other guys working out which was wrong so thank you for teaching me something new. Even thought the guys method was wrong would he get the marks as the answer was correct.
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3317752
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(Original post by Balkaran)
I didnt really know how to do it. I just went off the other guys working out which was wrong so thank you for teaching me something new. Even thought the guys method was wrong would he get the marks as the answer was correct.
the methods i used would still get u all the marks. I could not type out every single line of working properly from a computer near midnight.
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3317752
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(Original post by Balkaran)
That would be wrong I think
 -2 [(t-1.5)^2 - 2.75] = 0
You have to multiply - 2.75 by - 2
Giving you this answer
 -2(t-1.5)^2 +5.5 = 0
Then you take the 5.5 over
 -2(t-1.5)^2 =  -5.5
Divide it by -2
 -(t-1.5)^2 = 2.75
Making the points
 t = 1.5 h = 2.75
whats wrong? I got the right answer
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RogerOxon
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(Original post by Pretish)
whats wrong? I got the right answer
The equating to zero and your -5.5 (it's +5.5).
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mathcool
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Hello,I was looking for 9-1 GCSE maths and accidentally arrived here.Interger123 approach is right. Equating the quadratic expression (i.e. -2t squared plus six t 1) to zero is incorrect approach because what you are essentially saying or attempting to calculate by this approach is that value of time, t, when height above the ground is zero!. Yes, quadratic expression (i.e. -2t squared plus six t 1) to zero gives the times when rocket's height above the gound is zero, i.e. when the rocket is on the ground.If you think about it height of the rocket is given by the quadratic expression in terms of t (i.e time). So, you can verify the above at when t = 0.So, when t = 0 (by substituting zero for t in the quadratic expression), h = 1, this mean rocket was lauched 1 unit (what ever the unit, say 1 meter) from ground level. As you probably already know, the height of the rocket is traversing a prabolic profile (i.e. given the quadratic expression).What you have to use is calculas, i.e rate of change of height (i.e dh/dt = -4t 6), which I hope you have learnt it. At the time the rocket reaches maximum height, the rate of change of height (i.e. tangent to parbola will have zero gradient) at the point in time will be zero.Therefore, -4t 6 = 0, i,e t = 3/2 = 1.5. Even if the anser figure is correct (this is pure coicidence), the approach is completely wrong and you will probably receive zero marks because you have not shown (indeed shown incorrect understanding) the understanding and connection to physical meaning (i.e. problem's) and mathematical connection to the physical conditions and have not understood the physical significance of equating the quadratic expression (i.e. -2t squared plus six t 1) to zero.However, there is a way for using the equating the quadratic expression (i.e. -2t squared plus six t 1) to zero for gettting maximum height above the ground of the rocket. It is using the mathematical fact that parabolic profile's maximum height (i.e. vertex) will be on the middle point between the points where the quadratic expression (i.e. -2t squared plus six t 1) is zero. This is purely a mathematical approach.I hope my explanation helps you to connect the mathematical expression to physical conditions (i.e. in this case rocket's height above the ground).
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3317752
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(Original post by RogerOxon)
The equating to zero and your -5.5 (it's +5.5).
(Original post by Pretish)
whats wrong? I got the right answer
(Original post by Balkaran)
I didnt really know how to do it. I just went off the other guys working out which was wrong so thank you for teaching me something new. Even thought the guys method was wrong would he get the marks as the answer was correct.
(Original post by Integer123)
When you find the minimum or maximum (vertex) of a quadratic, you don't set it equal to zero, as it isn't being solved. You complete the square, as was done correctly, giving

h = -2(t-1.5)^2 + 5.5

It may help to write this as h = 5.5 -2(t-1.5)^2

Now, the maximum value of h is achieved when the smallest value of 2(t-1.5)^2 is subtracted from 5.5. As the bracket is squared, the smallest value to be subtracted is zero. Hence the maximum of h is 5.5 and it is achieved when t = 1.5.
(Original post by mathcool)
Hello,I was looking for 9-1 GCSE maths and accidentally arrived here.Interger123 approach is right. Equating the quadratic expression (i.e. -2t squared plus six t 1) to zero is incorrect approach because what you are essentially saying or attempting to calculate by this approach is that value of time, t, when height above the ground is zero!. Yes, quadratic expression (i.e. -2t squared plus six t 1) to zero gives the times when rocket's height above the gound is zero, i.e. when the rocket is on the ground.If you think about it height of the rocket is given by the quadratic expression in terms of t (i.e time). So, you can verify the above at when t = 0.So, when t = 0 (by substituting zero for t in the quadratic expression), h = 1, this mean rocket was lauched 1 unit (what ever the unit, say 1 meter) from ground level. As you probably already know, the height of the rocket is traversing a prabolic profile (i.e. given the quadratic expression).What you have to use is calculas, i.e rate of change of height (i.e dh/dt = -4t 6), which I hope you have learnt it. At the time the rocket reaches maximum height, the rate of change of height (i.e. tangent to parbola will have zero gradient) at the point in time will be zero.Therefore, -4t 6 = 0, i,e t = 3/2 = 1.5. Even if the anser figure is correct (this is pure coicidence), the approach is completely wrong and you will probably receive zero marks because you have not shown (indeed shown incorrect understanding) the understanding and connection to physical meaning (i.e. problem's) and mathematical connection to the physical conditions and have not understood the physical significance of equating the quadratic expression (i.e. -2t squared plus six t 1) to zero.However, there is a way for using the equating the quadratic expression (i.e. -2t squared plus six t 1) to zero for gettting maximum height above the ground of the rocket. It is using the mathematical fact that parabolic profile's maximum height (i.e. vertex) will be on the middle point between the points where the quadratic expression (i.e. -2t squared plus six t 1) is zero. This is purely a mathematical approach.I hope my explanation helps you to connect the mathematical expression to physical conditions (i.e. in this case rocket's height above the ground).

oh my life guys, the exact question was asked on a thread so i answered it there, and i just copied it. There are 2 ways to do this question, calculus and complting the square. In calculas u set the equation to 0, thats why i got confused and did the same here. Jeez people. XD
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RogerOxon
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(Original post by Pretish)
There are 2 ways to do this question, calculus and complting the square. In calculas u set the equation to 0,
If you use calculus, you have to differentiate h w.r.t. t, and set that equal to 0. It's quite important to correct any mistakes quickly, so that the OP (and others) don't learn an incorrect method. It's not a reprimand, it's purely making sure that the way to solve a problem is full understood.
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3317752
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#13
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(Original post by RogerOxon)
If you use calculus, you have to differentiate h w.r.t. t, and set that equal to 0. It's quite important to correct any mistakes quickly, so that the OP (and others) don't learn an incorrect method. It's not a reprimand, it's purely making sure that the way to solve a problem is full understood.
ik. I did differentiate h and t lol.
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