f.4 Maths !

[匿名]

2012-09-17 4:48 am

Quadratic Equations

1.The true bearing and compass bearing of a ship from a lighthouse are x(x+2)degree and S (3x+2)degree W respectively. Find the value of x.
ANS 14

2.Given that r is a root of the quadratic equation x^2-3x-5=0,find
the value of r^5 –3r^4-5r^3 +2r^2-6r .
ANS 10

3.The height and the perimeter of a right angled triangle are x cm and 40cm respectively if the length of its base is 1cm less than twice its height
find the value of x.
ANS 8

4.ABCD is a square of side 8 cm, P and Q are points on AB and BC respectively such that AP= QC =xcm. If the area of triangle ADP is the two third s of the area of triangle BPQ, find the value of x correct to 3 significant fig.
ANS2.51

5. Tom and Mary are due south and due east of a bus stop respectively, and the distance between them is 60m. They walk towards the bus stop at the same time with their own constant speeds.After 6 second, they are both 27m away from the bus stop. If Mary walks slower than Tom by2m/s, find their speeds.
ANS tom 3.5m/s Mary 1.5m/s

6.If (x-3)(x-4)=(a-3)(a-4) thenx=?
ANS a or 7-a

7a Solve 10x^2+9x-22=0
b. Mr tung deposited $10000 in abank on his 25th birthday and$9000on his 26 th birthday. The interest was compound yearly at r% p.a. and the total amount he received on his 27th birthday was $22000 Find r.
ANS a) –2 or 1.1 b)10

Please solve the above problems with full steps plz ((really wanna know the methods to do them ,,you may also add some explanation as well^^
Thanks so much<3

回答 (1)

用戶10012

2012-09-19 5:16 am

✔ 最佳答案

(1) Answered.
(2) Since r is the root of the equation, so r^2 - 3r - 5 = 0, that is r^2 - 3r = 5.
The expression = r^3(r^2 - 3r) - 5r^3 + 2(r^2 - 3r) = 5r^3 - 5r^3 + 2(5) = 10.
(3) Length of base = 2x - 1. By Pythagoras thm., hypotenuse = sqrt [ x^2 + (2x - 1)^2]
Perimeter = 40 = sqrt [ x^2 + (2x - 1)^2] + x + (2x - 1)
(41 - 3x)^2 = x^2 + 4x^2 + 1 - 4x
1681 - 246x + 9x^2 = 5x^2 - 4x + 1
4x^2 - 242x + 1680 = 0
2x^2 - 121x + 840 = 0
(x - 8)(2x - 105) = 0
x = 52.5 (rej.) or 8.

2012-09-18 21:29:57 補充：
(4) Area of ADP = 4x. Area of BPQ = (1/2)(8 - x)^2
so 4x = (1/3)(8 - x)^2.
12x = 64 - 16x + x^2
x^2 - 28x + 64 = 0
so x = 2.51 or 25.5 (rej.)

2012-09-18 21:42:23 補充：
(5)Let speed of Mary = v, so speed of Tom = v + 2
Distance traveled by Mary in 6 sec. = 6v
Distance traveled by Tom in 6 sec. = 6(v + 2)
So by Pythagoras thm., (27 + 6v)^2 + [ 27 + 6(v + 2)]^2 = 60^2
(2v - 3)(36v + 450) = 0, so v = 3/2 = 1.5 = speed of Mary, speed of Tom = 3.5.

2012-09-18 21:45:20 補充：
(6) (x - 3)(x - 4) = (a - 3)(a - 4).
Obviously, x = a is one of the 2 solution.
(x - 3)(x - 4) - (a - 3)(a - 4) = 0
x^2 - 7x + 12 - (a - 3)(a - 4) = 0
so sum of roots = 7
so the other root = 7 - a.

2012-09-18 21:52:15 補充：
(7) (a) you can handle.
(b) [10000(1 + R) + 9000](1 + R) = 22000
10000(1 + R)^2 + 9000(1 + R) - 22000 = 0
10(1 + R)^2 + 9(1 + R) - 22 = 0
From result of part (a) (1 + R) = - 2 or 1.1
so R = - 3 (rej) or 0.1 = 10%
so r = 10.

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