Question

Please help!
The equation x^3+px+q=0 has a repeated root. Prove that 4p^3+27q^2=0.

Answer 1

if given a polynomial degree's 3 :
ax^3 + bx^2 + cx + d = 0 (a,b,c, and d are contants)
then : some formula can to solve it
x1+x2+x3 = -b/a
x1x2 + x1x3 + x2x3 = c/a
x1 x2 x3 = -d/a

ur equation :
x^3+px+q=0
it means, a=4, b=0, c=p, d=q

has a repeated root, it means x1=x2=x3
now, we use the formulas above
x1+x2+x3 = -b/a
x1+x1+x1= 0
3x1 = 0
x1=0
thus x2=0 and x3=0

subtitute x=0 to the original equation :
x^3+px+q=0
0^3+p(0)+q=0
q=0
so, x^3+px = 0

next, use the formula :
x1x2 + x1x3 + x2x3 = c/a
3(x1)^2 = p
3(0)^2 = p
p=0

to prove 4p^3+27q^2=0
just put p=0 and q=0