中四數學三條問題

1a) x^4-17x^2+16
= (x^2 - 1)(x^2 - 16)
= (x^2 - 1)(x^2 - 4^2)
= (x - 1)(x + 1)(x - 2)(x + 2)
b) 2x^4-30x^2+20x+48
Let f(x) = 2(x^4 - 15x^2 + 10x + 24)
f(- 1) = 2(1 - 15 - 10 + 24) = 0 , so (x+1) is a factor ,
f(2) = 2(16 - 15(4) + 10(2) + 24) = 0 , so (x-2) is a factor ,
f(3) = 2(81 - 15(9) + 10(3) + 24) = 0 , so (x-3) is a factor ,
f(- 4) = 2(256 - 15(16) + 10(-4) + 24) = 0 , so (x+4) is a factor.
So 2x^4-30x^2+20x+48 = (x+1)(x-2)(x-3)(x+4)

2)It is given that y varies partly as x and partly as 1/x^2. When X=1, y=5 and when x=3,y=5, find x when y=30
Let y = (k1)x + (k2)(1/x^2) ,
5 = (k1) + (k2)......(1)
5 = 3(k1) + (k2)(1/9)......(2)
(1)*3 - (2) :
15 - 5 = 3(k2) - (1/9)(k2)
(k2) = 10 / (3 - 1/9) = 45/13
By (1) , 5 = (k1) + 45/13
(k1) = 20/13
So y = (k1)x + (k2)(1/x^2)
y = (20/13)x + (45/13)(1/x^2)
When y = 30 ,
30 = (20/13)x + (45/13)(1/x^2)
390 = 20x + 45/x^2
78 = 4x + 9/x^2
78x^2 = 4x^3 + 9
4x^3 - 78x^2 + 9 = 0
x = 19.49408 or - 0.33679 or 0.3427
I think your question have some typing error.


3)if 8sin^2x +6sinxcosx =2, find tanx
4sin^2x + 3sinxcosx = 1
(4sin^2x + 3sinxcosx) / cos^2x = (sin^2x + cos^2x) / cos^2x
4tan^2x + 3tanx = tan^2x + 1
3tan^2x + 3tanx - 1 = 0
tanx = {- 3 +/- √[(3^2 - 4(3)(-1)]} / (2*3)
tanx = (- 3 +/- √21)/6