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secondary 3 | A Maths
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need help with this question, pls explain too:)
Date Posted: 4 years ago
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For (i), since the 3 roots are 1, k and k²,
x = 1
x-1 = 0
x= k
x-k = 0
x = k²
x-k² = 0
If we work backwards and combine them,
(x-1)(x-k)(x-k²) = 0
There may or may not be any constant in front of the brackets. However, given the coefficient of x³ is -1, we know that there should be a “-1” there.
Hence,
f(x) = -(x-1)(x-k)(x-k²)
Then, we find the zero of “x-2”.
x-2 = 0
x = 2
Using remainder theorem, when x=2, we get a reminder of -3. Therefore, we substitute x=2 and make f(2) equal to -3.
f(2) = -(2-1)(2-k)(2-k²)
-3 = -(1)(2-k)(2-k²)
Divide/Multiply both sides by -1,
3 = (1)(2-k)(2-k²)
(2-k)(2-k²) = 3
4-2k²-2k+k³ = 3
k³-2k²-2k+4 = 3
k³-2k²-2k+4-3 = 0
k³-2k²-2k+1 = 0 (shown)
x = 1
x-1 = 0
x= k
x-k = 0
x = k²
x-k² = 0
If we work backwards and combine them,
(x-1)(x-k)(x-k²) = 0
There may or may not be any constant in front of the brackets. However, given the coefficient of x³ is -1, we know that there should be a “-1” there.
Hence,
f(x) = -(x-1)(x-k)(x-k²)
Then, we find the zero of “x-2”.
x-2 = 0
x = 2
Using remainder theorem, when x=2, we get a reminder of -3. Therefore, we substitute x=2 and make f(2) equal to -3.
f(2) = -(2-1)(2-k)(2-k²)
-3 = -(1)(2-k)(2-k²)
Divide/Multiply both sides by -1,
3 = (1)(2-k)(2-k²)
(2-k)(2-k²) = 3
4-2k²-2k+k³ = 3
k³-2k²-2k+4 = 3
k³-2k²-2k+4-3 = 0
k³-2k²-2k+1 = 0 (shown)
thx :)
No prob.
For (ii), it’s just like a normal cubic equation, except that it’s in terms of ‘k’ instead of ‘x’.
First, we let f(k) = k³-2k²-2k+1.
To find the roots of f(k), we could use the function in our calculator to help us get the 1st root, which is -1.
This means that the 1st factor of f(k) is (k+1). For instance,
k = -1
k+1 = 0.
Presentation wise, we have to write:
By trial and error,
f(-1) = 0.
To get the quadratic factor, we divide f(k) by (k+1). It could also be done using synthetic division since the factor we are dividing is linear.
If we do it right, the quadratic factor is:
(k²-3k+1)
Hence,
(k+1)(k²-3k+1) = 0
The quadratic factor cannot be factorised using frame method/cross diagram as they are irrational. Therefore, we’ll use the quadratic formula where:
a=1, b=-3, c=1
If we substituted the respective values and simplified the expression, we should get:
(-3+√5)/2 and (-3-√5)/2
First, we let f(k) = k³-2k²-2k+1.
To find the roots of f(k), we could use the function in our calculator to help us get the 1st root, which is -1.
This means that the 1st factor of f(k) is (k+1). For instance,
k = -1
k+1 = 0.
Presentation wise, we have to write:
By trial and error,
f(-1) = 0.
To get the quadratic factor, we divide f(k) by (k+1). It could also be done using synthetic division since the factor we are dividing is linear.
If we do it right, the quadratic factor is:
(k²-3k+1)
Hence,
(k+1)(k²-3k+1) = 0
The quadratic factor cannot be factorised using frame method/cross diagram as they are irrational. Therefore, we’ll use the quadratic formula where:
a=1, b=-3, c=1
If we substituted the respective values and simplified the expression, we should get:
(-3+√5)/2 and (-3-√5)/2
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