# Maths really hard question gcse

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#3

you had

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

so u got

then complete the square of whats inside the

so you got

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

from this you can see that the points are

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

so u got

then complete the square of whats inside the

so you got

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

from this you can see that the points are

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#4

(Original post by

you had

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

so u got

then complete the square of whats inside the

so you got

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

from this you can see that the points are

**Pretish**)you had

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

so u got

then complete the square of whats inside the

so you got

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

from this you can see that the points are

You have to multiply by

Giving you this answer

Then you take the 5.5 over

Divide it by -2

Making the points

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#5

(Original post by

That would be wrong I think

You have to multiply by

Giving you this answer

Then you take the 5.5 over

Divide it by -2

Making the points

**Balkaran**)That would be wrong I think

You have to multiply by

Giving you this answer

Then you take the 5.5 over

Divide it by -2

Making the points

**Pretish**)

you had

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

so u got

then complete the square of whats inside the

so you got

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

from this you can see that the points are

It may help to write this as

Now, the maximum value of is achieved when the smallest value of is subtracted from . As the bracket is squared, the smallest value to be subtracted is zero. Hence the maximum of is and it is achieved when .

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#6

(Original post by

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

It may help to write this as

Now, the maximum value of is achieved when the smallest value of is subtracted from . As the bracket is squared, the smallest value to be subtracted is zero. Hence the maximum of is and it is achieved when .

**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

It may help to write this as

Now, the maximum value of is achieved when the smallest value of is subtracted from . As the bracket is squared, the smallest value to be subtracted is zero. Hence the maximum of is and it is achieved when .

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#7

(Original post by

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.

**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.

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#8

**Balkaran**)

That would be wrong I think

You have to multiply by

Giving you this answer

Then you take the 5.5 over

Divide it by -2

Making the points

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#9

(Original post by

whats wrong? I got the right answer

**Pretish**)whats wrong? I got the right answer

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#10

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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#11

(Original post by

The equating to zero and your -5.5 (it's +5.5).

**RogerOxon**)The equating to zero and your -5.5 (it's +5.5).

(Original post by

whats wrong? I got the right answer

**Pretish**)whats wrong? I got the right answer

**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.

**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

It may help to write this as

Now, the maximum value of is achieved when the smallest value of is subtracted from . As the bracket is squared, the smallest value to be subtracted is zero. Hence the maximum of is and it is achieved when .

(Original post by

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).

**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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#12

(Original post by

There are 2 ways to do this question, calculus and complting the square. In calculas u set the equation to 0,

**Pretish**)There are 2 ways to do this question, calculus and complting the square. In calculas u set the equation to 0,

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#13

(Original post by

If you use calculus, you have to differentiate w.r.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.

**RogerOxon**)If you use calculus, you have to differentiate w.r.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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