O-Level Additional Math

[A concerned mom]

Hi Jie Heng,


Sorry for the trouble and appreciate if you could further enlighten me on the answer
=(x + 2) (2x^2 + x - 2) => how did you manage to get this?


2)a) x^2 – 4 + 2x^3 + 4x^2
=2x^3 + 5x^2 - 4
=(x + 2) (2x^2 + x - 2)

[jieheng]

A concerned mom:

Hi Jie Heng,


Sorry for the trouble and appreciate if you could further enlighten me on the answer
=(x + 2) (2x^2 + x - 2) => how did you manage to get this?


2)a) x^2 – 4 + 2x^3 + 4x^2
=2x^3 + 5x^2 - 4
=(x + 2) (2x^2 + x - 2)

Let f(x) = 2x^3 + 5x^2 - 4 = (ax + b) (Ax^2 + Bx + C)

i)To find a linear factor (ax + b)

The constant term of [f(x) = 2x^3 + 5x^2 - 4] is 4 and the factors of 4 are 1 , 2 and 4

We need to test whether x= 1, -1 , 2 , -2 , 4 and -4 are the solutions of f(x)=0

When x= 1 , f(1) = 2(1)^3 + 5(1)^2 -4 = 2 + 5 - 4 =3

When x= -1 , f(-1) = 2(-1)^3 + 5(-1)^2 -4 = -2 + 5 - 4 =-1

When x= 2 , f(2)= 2(2)^3 + 5(2)^2 -4 = 16 + 20 - 4 =32

When x= -2 , f(-2)= 2(-2)^3 + 5(-2)^2 -4 = -16 + 20 -4 = 0

As f(-2) = 0 , do not need to test f(4) and f(-4)
x= -2
x+2 = 0
(x+2) is a factor of f(x)


ii)To find a quadratic equation (Ax^2 + Bx + C)

f(x) = 2x^3 + 5x^2 - 4 = (x + 2) (Ax^2 + Bx + C)
2x^3 + 5x^2 - 4 = Ax^3 + BX^2 + Cx + 2Ax^2 + 2Bx + 2C
2x^3 + 5x^2 - 4 = Ax^3 + (2A+B)x^2 + (2B+C)x + 2C

Comparing the coefficients of x^3 , A =2

Comparing the coefficients of the constant term , 2C = -4 => C = -2

Comparing the coefficients of x^2 , 2A+B = 5 => 2(2) + B = 5 => B = 1

Ax^2 + Bx + C = 2x^2 + x -2

f(x) = 2x^3 + 5x^2 - 4 = (x + 2) (Ax^2 + Bx + C)
= (x + 2) (2x^2 + x -2)

This method will be taught in A Maths in Sec 3 ; Factor Theorem of Polynomials .

[A concerned mom]

Thank you Jie Heng for your great help!!^^

I think i need to study the syllabus of Sec 3 Add. Maths before analysing the answer.

[Valerie Teoh]

Any tips on doing A Maths Plane Geometry proving questions?

[JadeDry]

I tried solving these questions to no avail, could you please help me?



PQRS is a trapezium where PQ is parallel to SR. Given that the angle PQR = SPR = 90 degrees, PQ = 8.5 and PR = 12.3 calculate

A) PS

B) The area of the trapezium PQRS

And the second question-

ABCD is a trapezium, the angle ABC = the angle BCD = 90 degrees. Given that AB = 8 and, CD = 15, AD = 27 and BD = (3x - 4), calculate the value of X, giving your answer correct to two decimal places.

Thanks in advance!

[mom-of-2 0]

Can any kind soul please help me with this question please :


Find the equation of the circle which passes through the points A(5,-3), B(0,6) and has its centre lying on the line 2x-3y-6=0

Thank you very much in advance !

[WinnersEducation]

Concept:

1)Perpendicular bisector of any chord passes through the centre of circle.
2) Equation of circle : (x-a)^2 + (y-b)^2 = r^2 where (x, y) is any point on circumference. (a,b) : centre of circle, r: radius.

To solve this qn I would suggest drawing a diagram.

We need centre and radius of circle.

How to find centre?
- By solving simultaneous eqn of the given line and the perpendicular bisector or chord AB. Why? Both lines pass through centre.

- How to find eqn of perpendicular bisector of AB? We need gradient and a point on this perpendicular bisector.

Apply gradient formula on point A and B.
Relationship between AB and perpendicular bisector is m1.m2=-1. m1: gradient of AB. m2: gradient of perpendicular bisector of AB.

Then apply midpoint formula to find midpoint of AB. Why? Only this midpoint lies on the perpendicular bisector of AB.
Hence eqn of perpendicular bisector is formed.

Centre can be obtained.

How to find radius?
Apply distance formula between coordinate of centre (found earlier) and Point A or B (preferred since coordinate of B is 'prettier)'.

mom-of-2:

Can any kind soul please help me with this question please :

Find the equation of the circle which passes through the points A(5,-3), B(0,6) and has its centre lying on the line 2x-3y-6=0

Thank you very much in advance !

[mom-of-2 0]

Thank you very much !!

[jieheng]

JadeDry:

I tried solving these questions to no avail, could you please help me?



PQRS is a trapezium where PQ is parallel to SR. Given that the angle PQR = SPR = 90 degrees, PQ = 8.5 and PR = 12.3 calculate

A) PS

B) The area of the trapezium PQRS


Thanks in advance!

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[JadeDry]

Thanks for the help jieheng!


If you could help me with another question, I would appreciate it.

There is a hemisphere with a radius of 4cm. Some liquid is poured into the hemisphere to a depth of x cm. Show that the volume of the liquid is more than ( x cubed / 64) of the volume of the hemisphere.

Thanks in advance!