Completing a Cube

guitarguy

Junior Member
Joined
Feb 25, 2012
Messages
58
I am familiar with completing the square of a binomial.

I take the equation ax^2 + bx + c = 0
Subtract c from both sides and divide both sides by a
I then add b^2/(4*a^2) to both sides
Take the square root and I get the quadratic equation.

I want to derive a cubic equation for:

ax^3 + bx^2 + cx + d = 0

I subtracted d from both sides
divided both sides by a
Then I added (b^3*c^3)/(27*a^3)

This did not yield the solution of

(x + something)^3 = something

so I couldn't take the cube root and solve.

Any ideas?
 
I am familiar with completing the square of a binomial.

I want to derive a cubic equation for:

ax^3 + bx^2 + cx + d = 0
The formula you're looking for may be a bit more complicated that you're thinking it'll be. Look here. ;)
 
I am familiar with completing the square of a binomial.

I take the equation ax^2 + bx + c = 0
Subtract c from both sides and divide both sides by a
I then add b^2/(4*a^2) to both sides
Take the square root and I get the quadratic equation.

I want to derive a cubic equation for:

ax^3 + bx^2 + cx + d = 0

I subtracted d from both sides
divided both sides by a
Then I added (b^3*c^3)/(27*a^3)

This did not yield the solution of

(x + something)^3 = something

so I couldn't take the cube root and solve.

Any ideas?
\(\displaystyle \displaystyle (x + r)^3 = x^3 + 3x^2r + 3xr^2 + r^3\)

Derive r from the ratio first two coefficients:

\(\displaystyle b = 3ar\), or \(\displaystyle r = (1/3) b/a\)

[this corresponds exactly to \(\displaystyle r = (1/2)b/a\) for the "complete the square" case]

Then see what you have to add to both coefficients c and d to complete the cube.
 
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