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Urgent Help - OCR C3

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Original post by phat-chewbacca
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I can't finish the question. I have got up to mx^2 +(m^2+2)x +m = 0
Reply 2
All fine so far (although hard to read because your picture is sideways).

You are trying to show that there are exactly two stationary points. So you need the quadratic you have obtained to have certain properties. How can you check that it has the properties you want? If you have a go at this, you'll find that they've set the question up so that it is easy to do it.
Original post by Pangol
All fine so far (although hard to read because your picture is sideways).

You are trying to show that there are exactly two stationary points. So you need the quadratic you have obtained to have certain properties. How can you check that it has the properties you want? If you have a go at this, you'll find that they've set the question up so that it is easy to do it.


So should I solve the quadratic that I have obtained to determine the coordinates of the stationary points. But how will that work because when I solve for x, I will give me values of x in terms of m, right?
Original post by phat-chewbacca
So should I solve the quadratic that I have obtained to determine the coordinates of the stationary points. But how will that work because when I solve for x, I will give me values of x in terms of m, right?


You don't need to solve it. You simply need to show that mR\forall m\in \mathbb{R} the discriminant of the quadratic is greater than 0 as then there are 2 distinct solutions hence 2 stationary points.
(edited 7 years ago)
Original post by RDKGames
You don't need to show it. You simply need to show that for all mRm\in \mathbb{R} the discriminant of the quadratic is greater than 0 as then there are 2 distinct solutions hence 2 stationary points.


I find that hard to understand. I thought the discriminant if greater than will determine if the quadratic has two real roots on the x-axis (2 distinct roots). But how does that mean 2 stationary points?
Original post by phat-chewbacca
I find that hard to understand. I thought the discriminant if greater than will determine if the quadratic has two real roots on the x-axis (2 distinct roots). But how does that mean 2 stationary points?


You found dydx=emx(mx2+(m2+2)x+m)\frac{dy}{dx}=e^{mx}(mx^2+(m^2+2)x+m) and the stationary points are given by solving dydx=0\frac{dy}{dx}=0 so since emx0,mRe^{mx} \not=0, \forall m \in \mathbb{R} then the stationary points are given by the real solutions to the quadratic mx2+(m2+2)x+m=0mx^2+(m^2+2)x+m=0.

However you do not need to solve this quadratic as the question simply asks you show the amount of solutions for this quadratic. Now you find the amount of solutions to a quadratic by investigating the discriminant. If the discriminant of this quadratic is strictly greater than 0, then there are always 2 distinct solutions to the quadratic - hence dydx=0\frac{dy}{dx}=0 at 2 different points, thus 2 different stationary points.

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