Polynomials Question (1 Viewer)

[the-derivative]

Hey guys,

Wondering if anyone could help me with this question:

The equation x^3 + 3px^2 + 3qx + r = 0, where p^2 cannot = q has a double root. Show that (pq-r)^2 = 4(p^2-q)(q^2-pr).

I've tried finding P'(x) and letting it = 0, but i don't know where to go from there.

Thanks guys, in advance.

 

[Trebla]

Hey guys,

Wondering if anyone could help me with this question:

The equation x^3 + 3px^2 + 3qx + r = 0, where p^2 cannot = q has a double root. Show that (pq-r)^2 = 4(p^2-q)(q^2-pr).

I've tried finding P'(x) and letting it = 0, but i don't know where to go from there.

Thanks guys, in advance.

x³ + 3px² + 3qx + r = 0
3x² + 6px + 3q = 0
=> x² + 2px + p² = p² - q
=> (x + p)² = p² - q
x = - p ± √ (p² - q)
Sub x back into original equation and you should get your expression...

 

[the-derivative]

x³ + 3px² + 3qx + r = 0
3x² + 6px + 3q = 0
=> x² + 2px + p² = p² - q
=> (x + 2p)² = p² - q
x = - 2p ± √ (p² - q)
Sub x back into original equation and you should get your expression...

Oh ok, cool.
Thanks Trebla =)

 

[jet]

x³ + 3px² + 3qx + r = 0
3x² + 6px + 3q = 0
=> x² + 2px + p² = p² - q
=> (x + 2p)² = p² - q
x = - 2p ± √ (p² - q)
Sub x back into original equation and you should get your expression...

Shouldn't it be p, since (x + 2p)² = x² + 4px + 4p²
whilst (x + p)² = x² + 2px + p² like the line previous?

 

[Aerath]

Heh, Truong's question =P We just did it the other day. What Trebla said.

 

[exiting]

How many people from Truong's are actually in this forum?

 

[Trebla]

Shouldn't it be p, since (x + 2p)² = x² + 4px + 4p²
whilst (x + p)² = x² + 2px + p² like the line previous?

Yep, my bad. Thanks for pointing that out.