Polynomials - roots and coefficients (1 Viewer)

[Petinga]

1. if a and b are roots of the equation x^2 + 8x -5=0, find the quadratic equation whose roots are a/b and b/a.

2. If a and b are the roots of the equation x^2 + mx + n =0, find the roots of
nx^2 + (2n - m^2)x + n=0 in terms of a and b.

3. If a and b are roots of x^2 + 5x +7=0 form the quadratic equation whose roots are (a-b)^2 and (a+b)^2.

4. Solve the equation x^3 - 3x^2 - 4x +12 = 0 given that the sum of two of its roots is zero.

 

[KFunk]

4. Solve the equation x^3 - 3x^2 - 4x +12 = 0 given that the sum of two of its roots is zero.

Let the roots be a, b and c where a + b = 0. Using the sum of roots a + b + c = 3, ∴ c = 3 is a root. This means that (x-3) is a factor so you can divide by inspection or using polynomial division.

x³ - 3x² - 4x +12 = 0
(x - 3)(x² - 4) = 0
(x - 3)(x + 2)(x - 2)=0

hence the roots are 3, 2 and -2

 

[Trev]

1)
a+b=-b/a=-8
ab=c/a=-5
Roots a/b and b/a
Sum of roots = a/b+b/a = (a²+b²)/ab= [(a+b)²-2ab]/ab = [(-8)²-2.-5]/-5 = -74/5
Product of roots = 1
Equ = x²+(74/5)x+1

2)
x²+mx+n=0
a+b=-b/a=-m
ab=c/a=n
nx²+(2n-m²)x+n=0; sub m=-(a+b); n=ab into equation which gives:
abx²-(a²+b²)x+ab=0

3)
x²+5x+7=0
a+b=-5
ab=7
Roots (a-b)², (a+b)²
Sum of roots: 2(a²+b²) = 2[(a+b)²-2ab] = 2[(-5)²-2.7] = 22
Product of roots: (a²-b²)² = [(a+b)²-4ab]² = [(-5)²-4.7]² = 9
Equ: x²-22x+9=0

4)
x³-3x²-4x+12=0
Let roots be a,-a,b.
Roots (one at a time) = b = 3
Roots (two at a time) = -a² = -4; a=+/-2
Therefore x = 3, 2 or -2.

 

[robbo_145]

3)
Product of roots: (a²-b²)² = [(a+b)²-4ab]² = [(-5)²-4.7]² = 9
Equ: x²-22x+9=0

for product of roots
(a-b)²(a+b)²= 25(a-b)²
=25[(a+b)²-4ab]
= 25[25-4.7]
= -75

 

[Trev]

for product of roots
(a-b)²(a+b)²= 25(a-b)²
=25[(a+b)²-4ab]
= 25[25-4.7]
= -75

Bleh .....

 

[bubu]

kfunk why is a+b=0

 

[Trev]

a+b is zero, he has given a as one root and b as the negative root of the other so they cancel. Stated in the question was that the "sum of two of its roots is zero".