急 !急 ! F4 Quadratic Equations.

15a) (i) The tens digit = x - 3

(ii) Value of the no. = 10(x - 3) + x = 11x - 30

b) From the given, we have:

(x - 3)2 + x2 = 11x - 30 + 3

2x2 - 6x + 9 = 11x - 27

2x2 - 17x + 36 = 0

(2x - 9)(x - 4) = 0

x = 9/2 (rej.) or 4

So the no. is 14

16) Let Matthew is x yers old now, then Joan is (x - 2) years old and his fater is (x + 30) years old.

3 years ago, Matthew was (x - 3), Joan was (x - 5) and his fater was (x + 27)

Thus:

(x - 3)(x - 5) = 2(x + 27)

x2 - 8x + 15 = 2x + 54

x2 - 10x - 39 = 0

(x - 13)(x + 3) = 0

x = 13 or -3 (rej.)

So Matthew is 13 years old now.

18) Let BS = x, then AS = AR = 10 - x and SR = 2x

Using Pyth. thm:

2(10 - x)2 = (2x)2

2x2 - 40x + 200 = 4x2

x2 + 20x - 100 = 0

x = 10(√2 - 1) since x must be positive.

Then SA = 10(2 - √2)

Area of haxagon = 10 x 10 - [10(2 - √2)]2

= 65.7 cm2

19a) BE = 27 - 3x, BF = 9 - x

So area is (1/2)(27 - 3x)(9 - x)

= (3/2)(9 - x)(9 - x)

= (3/2)(9 - x)2

b) Area of AEH = 3x2/2

So:

(3/2)(9 - x)2 = 24x2

(9 - x)2 = 16x2

9 - x = 4x or 9 - x = -4x

x = 9/5 or -3 (rej.)

c) Area of EFGH = 27 x 9 - 3(9 - x)2 - 3x2

= 77.76 cm2