i need help (desperately) wid these maths sums.......plz help???

Question

1. Find all the common zeros of the polynomials y3 + 5y2 - 9y -45 and y3 + 8y2 +15y?
[numbers written after the variable 'y' are exponents ;i.e. 2 for square and 3 for cube]

2. Determine 'k' so that k+2 ,4k-6 and 3k-2 are three consecutive terms of an arithmetic progression?

3. lf H , C and V are respectively the height ,the curved surface area and the volume of a cone ,prove that :-
3(pi)VH3 [cube] -C2H2 [square] + 9v2 [square] = 0 ?

4. Given that one root of the quadratic equation ax2 + bx + c = 0 ,is three times the other root ,show that 3b2 = 16ac ?

5. The areas of three adjacent faces of a cuboid are x,y,z. lf the volume is v ,prove that v2 = xyz ?

thank u very much to all of u who have spent their precious time 2 answer my queries !!!


plz solve few more problems of mine..........plz

6. The base radii of 2 right circular cones of the same height are in the ratio 3:5.Find the ratio of their volumes?

7. Evaluate-
sin^2 5' + sin^ 10' + p +sin^ 85' + sin^ 90' =19/2 ? Find p?
[ ' symbol after the numeric values stands for degrees]

Answer 1

1)Lets solve y^3+8y^2+15y first.
If we have to find its zeros we have to equate this equation to 0
y^3+8y^2+15y=0
=>y(y^2+8y+15)=0
=>y(y^2+5y+3y+15)=0
=>y(y+5)(y+3)=0
=>y=0, y=-3, y=-5(These are zeroes of the polynomial)
Substitute all above values one at a time in polynomial y^3+5y^2-9y-45 you get this polynomial equal to 0 for every value.

2)If k+2, 4k-6 and 3k-2 are three consecutive terms of arithmetic progression, then all consecutive terms must differ by common difference d within the progression.
=>d=(4k-6)-(k+2) and d=(3k-2)-(4k-6)
=>d=4k-k-6-2 and d=3k-4k-2+6
=>d=3k-8 and d=-k+4
Since difference d is same,
d=3k-8=-k+4
=>3k-8=-k+4( only two considered at a time)
=>3k+k=8+4
=>4k=12
=>k=3

3)I couldn't prove this equation to be true. Seems that equation is not copied properly or I am not solving it correctly.

4)Assume x and y to be two roots for equation.
We have xy=c/a---(1) and x+y=-(b/a)---(2)
Also x=3y---(3)
Hence y^2=y.y=y(x/3)=(xy)/3---(4)
(x+y)^2=(-b/a)^2 (Squarring eq (2))
=>x^2+2xy+y^2=(b/a)^2(Expanding terms)
=>(3y)^2+2(3y)y+y^2=(b/a)^2---(eqn 3)
=>9y^2+6y^2+y^2=(b/a)^2
=>16y^2=(b/a)^2
=>{16(xy)}/3=(b/a)^2
=>16a^2(xy)=3(b^2)(Rearranging terms)
=>16a^2(c/a)=3(b^2) (Eqn 1)
=>16ac=3(b^2)
QED
5)Now if we consider the dimensions of cuboid as a,b,c then,
one of the area will be(say x), x=ab, the second(say y) y=bc, and the third(say z) z=ac
Hence R.H.S=xyz=(x)(y)(z)
=(ab)(bc)(ac)
=(a^2)(b^2)(c^2)
=(abc)^2
=V^2=L.H.S

Answer 2

For #2, a, b, c is an arithmetic progression if and only if b - a = c- b, which is to say 2b = a+c. Substitute in the three actual terms, and you'll have an easy equation to solve.

For #1, factor out y from the second polynomial and what's left is a quadratic term whose zeros are easy to find. Those two and 0 itself are the three zeros of the second one. Check whether any of those are also zeros of the first one.

Answer 3

(a million) hardship-loose roots of two polynoms are also roots of their massive difference 3y^2 +24 y +40 5 = 0 y=(-24 +-6)/6 = -3 and -5 . Now you ought to attempt each and each if???? thus because the 2d polyn has no indepndent time period you may write it as y(y^2+8y+15) with roots 0 ( not hardship-loose -3 and -5) attempt them contained in the first (2) The differnce between 2 consecutive words is the ratio so 4k-6 -ok-2 =r and 3k-2-4k+6=r 3k-8=r -ok+4 =r substracting 4k-12 =0 ok=3 r= a million Now keep on

Answer 4

For n1

Use "Euclidian Polynamial Division" Or Horner's Scheme..

You have to search, you cannot find it with mathematic calculus.

Example: Euclidian Division
y3+5y2-9y-45 |_y-1__
- (y3-y2)_____|y2+6y+3
6y2
- (6y2-6y)
3y
- (3y-3)
-42

etc...You risk to lost a lot of time with such techniques.

Look this one..
Factorization method
a)y3+5y2-9y-45=(y3-9y)+5(y2-9)=y(y-3)(y+3)+5(y-3)(y+3) => (y-3)(y+3)(y+5) => y={-5;-3;3}

I like this method :-)))))) Very rapid..
b)y3 + 8y2 +15y=y2+8y+15..................................It's primary class question lol...... y2+8y+15 => (y+5)(y+3) => y={-5;-3}

So the common zeros are {-5; -3}

With factorization, this question is very simple..

Answer 5

I will answer your math exercise number 4.

If one root of this equation is x and the other root is y, then we could say that y=3x.
So..

x+y=-b/a
x+3x=-b/a
4x=-b/a
x=-b/4a

Then....

xy=c/a
x3x=c/a
3x^2=c/a
3(-b/4a)^2=c/a
3b^2/16a^2=c/a
3ab^2=c16a^2
3ab^2/a=c16a^2/a(we can divide as a is not 0)
3b^2=16ac and that's the answer!

Answer 6

answer-1
the common zeroes are y3+y2
answer2