Question

find the equation whose roots are the squares of the roots of x^3+qx+r=0

Answer 1

Answer:

Let a,b,c be the roots of the cubic x^3 +qx+r=0;

then the roots of the required equation

are (b−c)^2 , (c−a)^2 , (a−b)^2

Also, a+b+c=0 , ab+bc+ac=q and abc= -r

Now (b−c)^2 = b^2+c^2−2bc

= a^2 + b^2 + c^2 − a^2 − (2abc)/a

=(a+b+c)^2 − 2(bc+ca+ab) - a^2 − (2abc)/a

= −2q − a^2 + (2r)/a;

Also when x = a in the given equation,

y = (b−c)^2 in the transformed equation;

∴ y = − 2q − x^2 + (2r)/x

Thus we have to eliminate x between the equations

x^3 + qx + r = 0 ,

and x^3 + (2q+y) x − 2r = 0

By subtraction (q+y)x = 3r ; or x = (3r)/ q+y

Substituting and reducing, we obtain

y^3+6qy^2+9q^2y+27r^2+4q^3 = 0

Answer 2

Answer:

x3+2qx2+r2=0

Step-by-step explanation:

I can't type that much so I just giving answer