How to solve this equation:#x^3+px^2+qx+r=0# knowing that #x_1/(p-q-r)=x_2/q=x_3/r#?

1 Answer
Cesareo R.
Jul 7, 2017

See below.

Explanation:

We have

#(x-x_1)(x-x_2)(x-x_3) = x^3-(x_1+x_2+x_3)x^2+(x_1 x_2+x_2 x_3+x_1 x_3)x-x_1x_2x_3#

and also

#x_1/(p-q-r)=x_2/q=x_3/r=lambda rArr {((x_1+x_2+x_3)/p = lambda), ((x_1 x_2+x_2 x_3+x_1 x_3)/((p-q-r)q+qr+(p-q-r)r)=lambda), ((x_1x_2x_3)/((p-q-r)q r)=lambda):}#

and then

#{(-p=lambda p),(q=lambda((p-q-r)q+qr+(p-q-r)r)),(-r=lambda(p-q-r)q r):}#

Here we have #lambda = -1#

Solving now

#{(q=-((p-q-r)q+qr+(p-q-r)r)),(1=(p-q-r)q):}#

now calling #mu = p-q-r# and solving

#{(q = -(mu q + q r + mu r)),(1 = mu q):}#

for #mu, r# we obtain

#{(mu=1/q), (r=-(q (1 + q))/(1 + q^2)):}#

and consequently

#{(x_1 = -1/q), (x_2 = -q), (x_3=(q (1 + q))/(1 + q^2)):}#

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