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junior college 1 | H1 Maths
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Date Posted: 3 years ago
Views: 420
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Can I write my answer like that? Or do I need to add k=0 and k-8=0, k=8
Date Posted:
3 years ago
Why have you written discriminant > 0?
Let y = x² + (k - 2)x + (k + 1)
When x² + (k - 2)x + (k + 1) > 0,
y > 0
So there is no intersection of the curve and the x-axis, where y = 0
Therefore, there are no real roots for the equation x² + (k - 2)x + (k + 1) = 0
Discriminant < 0
(k - 2)² - 4(1)(k+1) < 0
k² - 4k + 4 - 4k - 4 < 0
k² - 8k < 0
k(k - 8) < 0
k² - 8k is a upward sloping parabola (u-shaped).
If we let y = k² - 8k, it has an intersection with y = 0 (the k-axis) and will cut it at 2 points, where k = 0 and k = 8
Between k = 0 and k = 8, the curve is below the k-axis and so it will be negative.
So 0 < k < 8 for k(k - 8) < 0
Or
The other way to think about it:
① If k is between 0 and 8,
Then k is positive but k - 8 is negative. So their product would be negative, which is what we want.
② If k > 8 , both k and k-8 are positive. Their product is positive which does not satisfy our inequality.
③ Likewise, if k < 0 both k and k-8 are negative. Their product is positive which does not satisfy our inequality.
④ If k = 0 or k = 8, then one of the factors is 0. Then their product is 0 , which does not satisfy our inequality.
So 0 < k < 8
Set of values of k : ｛1,2,3,4,5,6,7｝
Or
｛ k : k ∈ R, 0 < k < 8｝
When x² + (k - 2)x + (k + 1) > 0,
y > 0
So there is no intersection of the curve and the x-axis, where y = 0
Therefore, there are no real roots for the equation x² + (k - 2)x + (k + 1) = 0
Discriminant < 0
(k - 2)² - 4(1)(k+1) < 0
k² - 4k + 4 - 4k - 4 < 0
k² - 8k < 0
k(k - 8) < 0
k² - 8k is a upward sloping parabola (u-shaped).
If we let y = k² - 8k, it has an intersection with y = 0 (the k-axis) and will cut it at 2 points, where k = 0 and k = 8
Between k = 0 and k = 8, the curve is below the k-axis and so it will be negative.
So 0 < k < 8 for k(k - 8) < 0
Or
The other way to think about it:
① If k is between 0 and 8,
Then k is positive but k - 8 is negative. So their product would be negative, which is what we want.
② If k > 8 , both k and k-8 are positive. Their product is positive which does not satisfy our inequality.
③ Likewise, if k < 0 both k and k-8 are negative. Their product is positive which does not satisfy our inequality.
④ If k = 0 or k = 8, then one of the factors is 0. Then their product is 0 , which does not satisfy our inequality.
So 0 < k < 8
Set of values of k : ｛1,2,3,4,5,6,7｝
Or
｛ k : k ∈ R, 0 < k < 8｝
Discriminant > 0 means there are two distinct real roots or 2 real solutions to the curve y = ax² + bx + c and x-axis (where y = 0). The curve cuts/intersects the axis at 2 distinct points.
Discriminant > 0 doesn't mean that y > 0.
If y > 0 , the y-value or y-coordinate of the curve is always bigger than 0.
This means that it does not intersect the x-axis at all (where y = 0) since all its y-values are above 0.
So there are no real roots and therefore discriminant < 0 when y > 0
Discriminant > 0 doesn't mean that y > 0.
If y > 0 , the y-value or y-coordinate of the curve is always bigger than 0.
This means that it does not intersect the x-axis at all (where y = 0) since all its y-values are above 0.
So there are no real roots and therefore discriminant < 0 when y > 0
done
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Do you mean the curve is like this, hence it does not touch the x-axis and hence it does not have any real roots, therefore b^2-4ac < 0?
Date Posted:
3 years ago
Exactly...
You can try the Desmos graphing calculator online
Plot y = x² + (k - 2)x + (k + 1)
There will be a slider function for you to adjust the value of k. Change it accordingly and see how the curve changes.
You'll notice that when you set k to anything between 0 and 8 (not inclusive), the curve never intersects the x-axis.
Very good for visualisation.
To understand why b² - 4ac < 0, you can revisit my explanation of the quadratic formula in your earlier question.
Plot y = x² + (k - 2)x + (k + 1)
There will be a slider function for you to adjust the value of k. Change it accordingly and see how the curve changes.
You'll notice that when you set k to anything between 0 and 8 (not inclusive), the curve never intersects the x-axis.
Very good for visualisation.
To understand why b² - 4ac < 0, you can revisit my explanation of the quadratic formula in your earlier question.
That’s cool. Thanks!
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