New Add Maths Questions
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This is the 4th pattern on the sum and product of roots of the quadratic equation (alpha and beta) in the new 2008 add maths syllabus.
Perhaps you will like to do the questions. These questions are not very difficult. It is just to show some tricks that "O" level examiners used in setting the questions on the quadratic equations (alpha and beta).
Question 1
Given that a and b are the roots of the equation x^2 = x - 2,
show that a^4 = 2 - 3a
Question 2
Given that a and b are the roots of the equation x^2 = x +7,
show that 1/a = (a-1)/7
[Note : Students cannot use cross-multiply to do the proofing]
Thank you for your kind attention.
Regards,
ahm97sic
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nice, even i have problem solving.Mind if i post the answer pls? To help those who have problems
Solution for question 2
x^2 - x - 7 = 0
a + b = 1
ab = -7
a = -7 / b
1 / a = b / -7
b = 1 - a
1 / a = ( 1 - a ) / -7
1 / a = ( a - 1 ) / 7 ( solved )
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solved meh ? shown lah
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Originally posted by Lee012lee:
solved meh ? shown lah
sama sama.
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I can't get question 1, only managed to solve for question 2. ):
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Originally posted by Ahm97sic:
This is the 4th pattern on the sum and product of roots of the quadratic equation (alpha and beta) in the new 2008 add maths syllabus.
Perhaps you will like to do the questions. These questions are not very difficult. It is just to show some tricks that "O" level examiners used in setting the questions on the quadratic equations (alpha and beta).
Question 1
Given that a and b are the roots of the equation x^2 = x - 2,
show that a^4 = 2 - 3a
Question 2
Given that a and b are the roots of the equation x^2 = x +7,
show that 1/a = (a-1)/7
[Note : Students cannot use cross-multiply to do the proofing]
Thank you for your kind attention.
Regards,
ahm97sic
Hi you mean solving the questions in the following manner is not allowed?
1) since a is a root, thus x =a
a^2 = a - 2
Squaring both sides
a^4 = a^2 - 4a + 4
a^4 = (a-2) - 4a +4
a^4 = 2-3a (shown)
If so, why is that so? Where is the mathematical flaw in the solution? As I understand, MOE's instructions to school teachers according to teacher friends of mine is that all mathematically correct solutions are to be accepted no?
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Hi Mikethm, crimsontactics and iamapebble
Question 1
Given that a and b are the roots of the equation x^2 = x - 2,
show that a^4 = 2 - 3a
Answer
Step 1 : Subst x = a into the equation
Since a is the root of the equation ie x = a, so we can substitute x = a into the
equation ie
x^2 = x - 2
a^2 = a - 2
Step 2 : To do the proofing ie use the LHS to prove until it is equal to the RHS or use the RHS to prove until it is equal to the LHS
LHS = a^4
= (a^2)^2
= (a-2)^2 since a^2 = a - 2
= a^2 -4a + 4
= a - 2 - 4a + 4 since a^2 = a - 2
= 2 - 3a
= RHS (Shown)
Now, you know the tricks that the "O" level examiners used to set questions on this 4th pattern of the quadratic equation (alpha and beta) ie first substitute x = a into the equation, next use this result or some re-arranging of the result in the proofing.
Thank you for your kind attention.
Regards,
ahm97sic
PS : Dear Mikethm, your answer is actually correct. It is just that you have not presented your answer in the proper way ie for proofing, we use the LHS to prove until it is equal to the RHS or use the RHS to prove until it is equal to the LHS.
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Originally posted by Ahm97sic:
PS : Dear Mikethm, your answer is actually correct. It is just that you have not presented your answer in the proper way ie for proofing, we use the LHS to prove until it is equal to the RHS or use the RHS to prove until it is equal to the LHS.
Actually personally I am quite fussy about "properness" but I make sacrifices in presentation standard to make it understandable and duplicatible for my students. The proper LHS====>RHS presentation is what I prefer too... but doing it the way I do by starting out with the original equation is far easier for my students it seem. So I will have to live with it.
Likewise, for geometric proof, the so called proper presentation of statement/reason (US high school style) is totally not used by school teachers too. And I ain't going to complain coz it is easier not having to present it that way.
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sry man could you explain why x=a?
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Because it "a" is the root of the equation.
Need an example?
The roots of an equation, x^2 + 2x - 3 = 0, are -3 and 1, right?
So i can sub -3 or 1 into x^2 + 2x - 3=0
Let f(x) = x^2 + 2x - 3
so f(-3) = 9 - 6 - 3 = 0 (proved)
f(1) = 1 + 2 - 3 = 0
so if a = -3 and b = 1 (i said IF), i can sub "a" or "b" into the equation of the question right? Since the question requires proving of a^4..., subbing "b" is redundant as there is no "b" to be proved.
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Originally posted by Garrick_3658:
Because it "a" is the root of the equation.
Need an example?
The roots of an equation, x^2 + 2x - 3 = 0, are -3 and 1, right?
So i can sub -3 or 1 into x^2 + 2x - 3=0
Let f(x) = x^2 + 2x - 3
so f(-3) = 9 - 6 - 3 = 0 (proved)
f(1) = 1 + 2 - 3 = 0
so if a = -3 and b = 1 (i said IF), i can sub "a" or "b" into the equation of the question right? Since the question requires proving of a^4..., subbing "b" is redundant as there is no "b" to be proved.
Helped alot in clearing the concept. thanks