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junior college 2 | H2 Maths
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Pls help me with both parts. Thanks
Date Posted: 4 years ago
Views: 517
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I am very rusty in A Level Maths as I have not touched them for years even though I attempted a few A Level Maths questions in this forum so far.
Therefore, I am unable to guarantee that my method is valid, but I have rechecked my solutions against the two equations and found that my solutions are valid when placed back into the equations.
My approach in this question is to solve this question the way I would for an O Level (or similar level) polynomial equation.
That is, to guess a factor (except that the factor is already given), perform long division or compare coefficients and then factorise the polynomial expression, except that this time i is involved.
The resultant quadratic equation can be easily solved using the quadratic formula, bearing in mind that i2 = -1.
Finally, the two "polynomial" equations given are similar looking except that one contains all plus signs and the other alternates between pluses and minuses.
If you have done binomial expansions long enough in O Level (or similar level), you would have noticed that replacing the second term by the negative of itself will transform the first form into the second form. For example, (x + 1)^3 = x3 + 3x2 + 3x + 1 while (x - 1)^3 = x3 - 3x2 + 3x - 1.
A similar idea works here, so by inspection, letting w = -z transforms the equation in w to the equation in z. Hence, the roots w are equivalent to the negatives of all the roots of z.
Therefore, I am unable to guarantee that my method is valid, but I have rechecked my solutions against the two equations and found that my solutions are valid when placed back into the equations.
My approach in this question is to solve this question the way I would for an O Level (or similar level) polynomial equation.
That is, to guess a factor (except that the factor is already given), perform long division or compare coefficients and then factorise the polynomial expression, except that this time i is involved.
The resultant quadratic equation can be easily solved using the quadratic formula, bearing in mind that i2 = -1.
Finally, the two "polynomial" equations given are similar looking except that one contains all plus signs and the other alternates between pluses and minuses.
If you have done binomial expansions long enough in O Level (or similar level), you would have noticed that replacing the second term by the negative of itself will transform the first form into the second form. For example, (x + 1)^3 = x3 + 3x2 + 3x + 1 while (x - 1)^3 = x3 - 3x2 + 3x - 1.
A similar idea works here, so by inspection, letting w = -z transforms the equation in w to the equation in z. Hence, the roots w are equivalent to the negatives of all the roots of z.
Date Posted:
4 years ago
Thanks so much! I checked your answers and they are correct.
w³ - 2iw² + 5w - 6i = 0
The easiest trick here is since you have -6i here, change it to +6i to match the equation first.
You can do this by multiplying the entire equation by -1. The right hand side remains as 0 since -1 x 0 is still 0.
Or, you can bring all the left hamd side terms to the right hand side.
so -w³ + 2iw² - 5w + 6i = 0
From here it is easy to rewrite it in the form of the first equation.
(-w)³ + 2i(-w)² +5(-w) + 6i = 0
Comparing both equations' variables,
z = -w
so w = -z
As for the quadratic equation z² + iz + 6 = 0, factorisation is actually quite simple without the use of quadratic formula.
As there is an i in the coefficient of z, the constant terms in your factors will contain i. So you just need to change your 6 into terms of i. Since i² = 1, -i² = 1. This comes in useful at times.
z² + iz + 6 = 0
z² + iz - 6i² = 0
(z - 2i)(z + 3i) = 0
z = 2i or z = -3i
This is analogous to z² + z - 6 = 0
where z = 2 or z = -3
You can put any coefficient for z and the coefficient² as a multiple of the constant term . Your z would = 2(coefficient) or -3(coefficient)
Eg. Let the coefficient be 10,
z² + 10z - 6(10²) = 0 would give you the result z = 10 x 2 = 20 or z = 10 x -3 = -30
The easiest trick here is since you have -6i here, change it to +6i to match the equation first.
You can do this by multiplying the entire equation by -1. The right hand side remains as 0 since -1 x 0 is still 0.
Or, you can bring all the left hamd side terms to the right hand side.
so -w³ + 2iw² - 5w + 6i = 0
From here it is easy to rewrite it in the form of the first equation.
(-w)³ + 2i(-w)² +5(-w) + 6i = 0
Comparing both equations' variables,
z = -w
so w = -z
As for the quadratic equation z² + iz + 6 = 0, factorisation is actually quite simple without the use of quadratic formula.
As there is an i in the coefficient of z, the constant terms in your factors will contain i. So you just need to change your 6 into terms of i. Since i² = 1, -i² = 1. This comes in useful at times.
z² + iz + 6 = 0
z² + iz - 6i² = 0
(z - 2i)(z + 3i) = 0
z = 2i or z = -3i
This is analogous to z² + z - 6 = 0
where z = 2 or z = -3
You can put any coefficient for z and the coefficient² as a multiple of the constant term . Your z would = 2(coefficient) or -3(coefficient)
Eg. Let the coefficient be 10,
z² + 10z - 6(10²) = 0 would give you the result z = 10 x 2 = 20 or z = 10 x -3 = -30
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