O-Level Additional Math

[S-H]

Thank you very much!! MathIzzzFun. :thankyou:

[pinky88]

Hi all, pls help on the following 2 qns:


1) If x & y are positive integers which satisfy x² - 8x - 1001y² = 0, what is
the smallest possible value of x + y?

2a) If y = |x| + |x + 1| + |x - 1|, find the minimum value of y.

b) Find the area, in square units, enclosed by the figure whose boundary
points satisfy |x|+ |y|=4

TIA.

[lost boy]

Pls help me to slove sec 3 emath


Factorise each of the following completely.
a) ab^2 - 16a
b) (x^4 - x^2) + (x^2 - 1)
c) m - n - 1 + mn

Thanks! !

[kiasu_student]

Hi there, although I am still not required to know cubic equations until few years time, but I am too kiasu, can ask for any simplified formula or must I use inequalities and quadratic formula?

[jieheng]

lost boy:

Pls help me to slove sec 3 emath


Factorise each of the following completely.
a) ab^2 - 16a
b) (x^4 - x^2) + (x^2 - 1)
c) m - n - 1 + mn

Thanks! !

a)

ab^2 - 16a
= a(b^2 - 16)
=a(b^2 - 4^2)
=a(b+4)(b-4)

b)

(x^4 - x^2) + (x^2 - 1)
=x^2 (x^2 - 1) + (x^2 - 1)
=(x^2 + 1)(x^2 - 1)
=(x^2 + 1)(x + 1)(x - 1)

c)

m - n - 1 + mn
=m(1 + n) - (n + 1)
=(m - 1) (n + 1)

[lost boy]

Thank u jieheng can u pls help me another ques


Solve the simultaneous equations.

2x - 5y= 12
3x - 2y = 7

Thanks

[jieheng]

lost boy:

Thank u jieheng can u pls help me another ques


Solve the simultaneous equations.

2x - 5y= 12
3x - 2y = 7

Thanks

2x - 5y= 12 -----(1)
3x - 2y = 7 -----(2)

(1) x3 , 6x - 15y = 36 => 6x = 36 + 15y

(2) x2 , 6x - 4y = 14 => 6x = 14 + 4y

36 + 15y = 14 + 4y
11y = - 22
y = -2

6x = 14 + 4y
6x = 14 + 4(-2)
x= 1

[lost boy]

Thank you Jieheng.

[lost boy]

Pls help me to solve sec 3 a math


Use suitable substituitions to solve each of the following equations.

a) 5^x + 5 = 30(5^x-1)

b) 9^x + 10(3^x) = 3^x+2 + 12

Thanks

[jieheng]

lost boy:

Pls help me to solve sec 3 a math


Use suitable substituitions to solve each of the following equations.

a) 5^x + 5 = 30(5^x-1)

b) 9^x + 10(3^x) = 3^x+2 + 12

Thanks

Pls note the difference between (5^x-1) = (5^x) - 1 and [5^(x-1)]


a)

5^x + 5 = 30[5^(x-1)]

5^x + 5 = 30[(5^x)/5]

5^x + 5 = 6(5^x)

5(5^x) = 5

5^x = 1

x= 0

b)

9^x + 10(3^x) = 3^(x+2) + 12

3^(2x) + 10(3^x) = 3^x * 9 + 12

3^(2x) + 3^x - 12 = 0

Let y = 3^x

y^2 + y -12 = 0

(y + 4)(y - 3) = 0

y = -4 (reject) or y = 3

when y =3

3^x = 3

x = 1