Fudge2
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Hi, I'm looking for a bit of help on question 4biii. I understand what has been done, but not why. You've factorised x^3-4x+15=0 earlier in the question, so the mark scheme just shows that the discrimant of the quadratic factor is less than 0 so has no real roots, making the only real root -3. What I don't understand is how the fact that -3 is the only real root means that it's the only stationary point. I was only told that a root is when the curve crosses the x axis!

Could anyone be awesome enough to help me out?
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Sir Cumference
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(Original post by Fudge2)
Hi, I'm looking for a bit of help on question 4biii. I understand what has been done, but not why. You've factorised x^3-4x+15=0 earlier in the question, so the mark scheme just shows that the discrimant of the quadratic factor is less than 0 so has no real roots, making the only real root -3. What I don't understand is how the fact that -3 is the only real root means that it's the only stationary point. I was only told that a root is when the curve crosses the x axis!

Could anyone be awesome enough to help me out?
In part bii, you've shown that the x-coordinates of any stationary points satisfy the equation

x^3-4x+15=0

So if there is only one real root of this equation (x=-3) then the stationary point of the curve must have x-coordinate -3.

Does that make sense?

I think you're trying to link roots to stationary points in a graphical way but you're forgetting part bii where the link has been made.
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WingedCurves
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(Original post by Fudge2)
Hi, I'm looking for a bit of help on question 4biii. I understand what has been done, but not why. You've factorised x^3-4x+15=0 earlier in the question, so the mark scheme just shows that the discrimant of the quadratic factor is less than 0 so has no real roots, making the only real root -3. What I don't understand is how the fact that -3 is the only real root means that it's the only stationary point. I was only told that a root is when the curve crosses the x axis!

Could anyone be awesome enough to help me out?
You have found  \frac{dy}{dx} of the function, which yields a cubic. In part 4aii you have factorized this cubic and have found that it only has one real root.

Recall that the solutions to the equation  \frac{dy}{dx} = 0 are the x coordinates of the stationary points of the curve. Since there is only one real solution to the cubic (and by association  \frac{dy}{dx} ) there is only one stationary point to the curve.

Does this make sense?
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Fudge2
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(Original post by notnek)
In part bii, you've shown that the x-coordinates of any stationary points satisfy the equation

x^3-4x+15=0

So if there is only one real root of this equation (x=-3) then the stationary point of the curve must have x-coordinate -3.

Does that make sense?

I think you're trying to link roots to stationary points in a graphical way but you're forgetting part bii where the link has been made.
(Original post by WingedCurves)
You have found  \frac{dy}{dx} of the function, which yields a cubic. In part 4aii you have factorized this cubic and have found that it only has one real root.

Recall that the solutions to the equation  \frac{dy}{dx} = 0 are the x coordinates of the stationary points of the curve. Since there is only one real solution to the cubic (and by association  \frac{dy}{dx} ) there is only one stationary point to the curve.

Does this make sense?
So is a root the same as a real solution? ie the only solution is -3, so therefore if the stationary point satisfies the equation it must have x co-ordinate -3? Or have I got the wrong end of the stick...xD

Thanks both!
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Mattematics
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(Original post by Fudge2)
So is a root the same as a real solution? ie the only solution is -3, so therefore if the stationary point satisfies the equation it must have x co-ordinate -3? Or have I got the wrong end of the stick...xD

Thanks both!
dy/dx allows you to find the gradient of a function at a given point. You differentiated the original equation to get equation two (the cubic).

At a stationary point, the gradient is 0 (as at the point of turning the tangent to the line would be horizontal). So, you've equalled the cubic equation (dy/dx, the gradient) to 0 and factorised. You've then found that the CUBIC (aka gradient) has only one root/solution.

Every real solution to the cubic would be an x co-ordinate where the gradient is 0 (a stationary point). As there's only one real solution to dy/dx=0 (the cubic) there's only one stationary point.
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Fudge2
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(Original post by Mattematics)
dy/dx allows you to find the gradient of a function at a given point. You differentiated the original equation to get equation two (the cubic).

At a stationary point, the gradient is 0 (as at the point of turning the tangent to the line would be horizontal). So, you've equalled the cubic equation (dy/dx, the gradient) to 0 and factorised. You've then found that the CUBIC (aka gradient) has only one root/solution.

Every real solution to the cubic would be an x co-ordinate where the gradient is 0 (a stationary point). As there's only one real solution to dy/dx=0 (the cubic) there's only one stationary point.
Cool. I think I get it!

Thanks for your help everyone.

Just as an added point, why do we say 'real solution' rather than 'solution'?


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WingedCurves
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(Original post by Fudge2)
So is a root the same as a real solution? ie the only solution is -3, so therefore if the stationary point satisfies the equation it must have x co-ordinate -3? Or have I got the wrong end of the stick...xD

Thanks both!
A root of a polynomial  f(x) is the same as a solution of the equation  f(x) = 0 . In this case, since x = -3 is the only real root of the cubic, it is therefore the only (real) solution to the equation (cubic) = 0. As the cubic also represents the gradient of the curve ( \frac{dy}{dx} ) and we know a stationary point is at  \frac{dy}{dx} = 0 , then the real solutions to (cubic) = 0 are the x coordinates of the stationary point.
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Mattematics
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(Original post by Fudge2)
Cool. I think I get it!

Thanks for your help everyone.

Just as an added point, why do we say 'real solution' rather than 'solution'?


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Ever heard of complex/imaginary numbers? As a brief rundown: the square root of -1 is represented by i. So for instance, if you were using the quadratic formula and had a negative square root, the solution would be a complex number.

A complex number has two components - a 'real part', e.g. 5, and an imaginary part, e.g. 5i.

A real number has no imaginary component - no i. So, we say 'real solutions' to distinguish from complex solutions.

That's further maths stuff though, so don't worry about it if you're not doing further/haven't done Further Pure modules.
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Fudge2
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(Original post by Mattematics)
Ever heard of complex/imaginary numbers? As a brief rundown: the square root of -1 is represented by i. So for instance, if you were using the quadratic formula and had a negative square root, the solution would be a complex number.

A complex number has two components - a 'real part', e.g. 5, and an imaginary part, e.g. 5i.

A real number has no imaginary component - no i. So, we say 'real solutions' to distinguish from complex solutions.

That's further maths stuff though, so don't worry about it if you're not doing further/haven't done Further Pure modules.
Ahh, the wonderful world of further maths. Thanks.
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WingedCurves
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(Original post by Fudge2)
Cool. I think I get it!

Thanks for your help everyone.

Just as an added point, why do we say 'real solution' rather than 'solution'?


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There are a set of numbers that are an extension of the 'real numbers' (denoted as  \mathbb{R} ) called the 'complex numbers' (denoted as  \mathbb{C} ) which may have to be used to solve certain equations such as this one. I can elaborate if you would like, however it is suffice to say that not all polynomials have solutions in the real numbers, and so these numbers must be used to completely solve them
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Fudge2
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(Original post by WingedCurves)
There are a set of numbers that are an extension of the 'real numbers' (denoted as  \mathbb{R} ) called the 'complex numbers' (denoted as  \mathbb{C} ) which may have to be used to solve certain equations such as this one. I can elaborate if you would like, however it is suffice to say that not all polynomials have solutions in the real numbers, and so these numbers must be used to completely solve them
Your explanation is enough - I don't do Further Maths so I can stay in the 'real' world
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swinroy
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(Original post by Fudge2)
Cool. I think I get it!

Thanks for your help everyone.

Just as an added point, why do we say 'real solution' rather than 'solution'?


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That’s a great question. It’s to distinguish it from Complex Solutions.
Consider this equation
X-squared = -1
This has TWO non- real solutions
X = i
X= -i
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RDKGames
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(Original post by swinroy)
That’s a great question. It’s to distinguish it from Complex Solutions.
Consider this equation
X-squared = -1
This has TWO non- real solutions
X = i
X= -i
Bit late to the party

:party:
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swinroy
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(Original post by RDKGames)
Bit late to the party

:party:
What’s five years between friends LOL
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