Polynomials (1 Viewer)

[Petinga]

help needed asap plz

1. If the roots of the equation x^3 + px^2 + qx + r=0 form consecutive terms of a geometric sequence, prove that q^3=p^3r.
Show that this condition is satisfied for the equation 8x^3 - 100x^2 + 250x -125=0 and solve this equation.

2. Find two values of m such that the roots of the equation x^3 + 2x^2 +mx -16=0 are alpha, beta, alpha beta and using these values of m find alpha and beta.

3. If the roots of the equation 24x^4 - 52x^3 +18x^2 +13x - 6=0 are alpha, - alpha, beta and 1/beta. Find alpha and beta.

4. If 2 of the roots of the equation x^3 + qx + r=0 are equal, show that 4q^3 + 27r^2=0.

5. If the roots of the equation x^3 + 3x^2 -2x + 1=0 are alpha, beta and gamma. Find the value of:
a)alpha^2(beta + gamma) + beta^2(gamma + alpha) + gamma^2(alpha + beta).
b)alpha^2beta^2 + beta^2gamma^2 + gamma^2alpha^2

 

[Trev]

1)
x³+px²+qx+r=0; let roots = a/r, a, ar
[one at a time]; a/r+a+ar = a(1/r+1+r) = -p
[two at a time]; a²/r+a²+a²r = a²(1/r+1+r) = q
[three at a time]; a³ = -r
(1/r+1+r) = -p/a = q/a²
-ap=q; since a=(-r)^1/3
-(-r)^1/3p=q and q³=p³r.

8x³-100x²+250x-125=0; x³-(100/8)x²+(250/8)x-(125/8)=0;
q=(25/8); p=(-100/8); r=(-125/8); sub into q³=p³r and to show condition is satisfied (equal).
Then use a³=-r, etc to solve equation.

 

[Trev]

2)
x³+2x²+mx-16=0; Let roots = a, b, ab.
a+b+ab=-2
ab+ab(a+b)=m
(ab)²=16; ab=+/-4.
Using m=ab+ab(a+b)
When ab=4; a+b=-6 so m=-20.
When ab=-4; a+b=2 so m=-12.
m=-20 or -12.
Equ. is x³+2x²-20x-16=0 or x³+2x²-12x-16=0
Then use simultanous equations to find values of a and b.

 

[Trev]

3)
[roots one at a time] = b+1/b=13/6
[roots two at a time] = 1+a² (the rest cancels) = 3/4
[roots three at a time] = -a²(b+1/b) = -13/24
[roots four at a time] = -a²=-1/4
Solve the quadratic b+1/b=13/6 to find values of b.
Roots are [a, -a] 1/2, -1/2, [b, 1/b] 2/3, 3/2.

 

[Trev]

x³+qx+r=0; 4q³+27r²=0.
Let roots = a, a, b.
[roots one at a time] = 2a+b=0; -2a=b (1)
[roots two at a time] = a²+2ab=q (2)
[roots three at a time] = a²b=-r (3)

(1) in (2) a²-4a²=q; q=-3a²
(1) in (3) -2a³=-r; a=(r/2)^1/3
So q=-3[(r/2)^1/3]²
q³=-27r²/4 and 4q³+27r²=0.

 

[Trev]

5a)
Let roots =a, b, y.
Find a²(b+y)+b²(a+y)+y²(a+b)

a+b+y=-3; so a+b=-y-3 ∴ b+y=-3-a and a+y=-b-3
a²(b+y)+b²(a+y)+y²(a+b)
= a²(-3-a)+b²(-3-b)+y²(-3-y)
= -3(a²+b²+y²)-(a³+b³+y³)

From x³+3x²-2x+1=0; let x=a as a root, so:
a³=-3a²+2a-1
a³+b³+y³ = -3(a²+b²+y²)+2(a+b+y)-3

-3(a²+b²+y²)-(a³+b³+y³) becomes:
-3(a²+b²+y²)-(-3(a²+b²+y²)+2(a+b+y)-3)
= -2(a+b+y)+3; since a+b+y=-3 then = -2.-3+3 = 9.
So, a²(b+y)+b²(a+y)+y²(a+b) = 9.

 

[Riviet]

*claps* You are one hell of a genius trev