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secondary 3 | A Maths
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How to do Qn 2 and 4b????
Qn 4b wants to find the value of p and k
Date Posted: 6 years ago
Views: 825
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Equate both equations to get the solutions. Since its a tangent, it only has one solution.
With the quadratic equation you get:
0= x^2 - 2k^2x + k^4
You can either use perfect square to find the solution, or use b^2 - 4ac = 0.
If you use b^2 - 4ac =0, you get
4k^4 - 4(1)(4k^4) = 0.
Since this equation is true regardless of any value of k, k is from negative infinity to positive infinity
Hope it helps!
With the quadratic equation you get:
0= x^2 - 2k^2x + k^4
You can either use perfect square to find the solution, or use b^2 - 4ac = 0.
If you use b^2 - 4ac =0, you get
4k^4 - 4(1)(4k^4) = 0.
Since this equation is true regardless of any value of k, k is from negative infinity to positive infinity
Hope it helps!
Date Posted:
6 years ago
Thx
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Since the equation is less than 0 from 4 to k, the solutions of the equation are 4 and k. From there you can find the values of k and p by substituting the appropiate values. Hope it helps!
Date Posted:
6 years ago
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