another polynomial question (1 Viewer)

[Jaydels]

I have just one more question:

If ax^3 +bx^2 +d =0 has a double root, show that 27a^2d =4b^3 =0

thanks....

 

[grimreaper]

P(x)= ax^3 + bx^2 + d = 0

P'(x)= 3ax^2 + 2bx = 0
x^2 + 2bx/3a = 0

complete the square, find an expression for x, sub it into ax^3 + bx^2 + d = 0

Theres an easier way than that im certain but havent done maths in over 2 months lol

 

[Jaydels]

oh thanks, i differentiated but didnt know where to go from there

 

[Euler]

P(x)= ax^3 + bx^2 + d = 0

P'(x)= 3ax^2 + 2bx = 0
x^2 + 2bx/3a = 0

complete the square, find an expression for x, sub it into ax^3 + bx^2 + d = 0

Theres an easier way than that im certain but havent done maths in over 2 months lol

just like u said...
seeing that 3ax^2 + 2bx = 0, then either x=0 or the other nonzero one (too lazy to type it out).

clearly, x=0 is not the double root, since P(0)=d (ok...assume d to be nonzero)
then the double root must be the other one.
so, P(other one)=0 and this gives you what you need to prove (typo in the first post).