Expert: Josh - 4/19/2007

Question
question 1: If a,b,c are the roots of the equation x^3-7x^2+x+5=0; find the equation whose roots are a^2+b^2, b^2+c^2, c^2+a^2.
question 2: find the equation of the line joining the points (7,9,-3) and (11,-5,-2).

Answer
Dear Arun,

Q1. If a,b,c are the roots of the equation x^3-7x^2+x+5=0, they will all satisfy the equation. So, we have
a^3-7a^2+a+5=0
b^3-7b^2+b+5=0
c^3-7c^2+c+5=0.
This is not very helpful. Instead, observe that the last coefficient must be the product of all the roots.

i.e., abc=5.

Now, pick x=1 (a trial solution). Substituting this into the equation, we find that x=1 satisfies the equation (1-7+1+5=0). So, a=1 must be one of the roots.

x^3-7x^2+x+5 can now be factorized as (x-1)P(x), where P(x) is a polynomial of degree 2. Using long division, we find that P(x)=(x^2-6x-5)=(x-b)(x-c), where b=3+sqrt(14), c=3-sqrt(14) using the quadratic formula.

If a cubic polynomial has the roots a^2+b^2, b^2+c^2, c^2+a^2, then, [x-(a^2+b^2)][x-(b^2+c^2)][x-(c^2+a^2)]=0.
Substitute the known values of a,b,c in the above expression. Expanding it, you'll get the polynomial.

Q2. I'm not sure if you have been shown this. The result comes form vector algebra. Let R=(7,9,-3), P=(11,-5,-2) and V=(x,y,z). A straight line in the X-Y-Z plane passing through R and P may be written in the vector parametric form as:

V=R+t(P-R). viz., (x,y,z)=(7,9,-3)+t(11-7,-5-9,-2+3)