+0

# Solve: x^3 + (p+x)x^2 +2px = 0

0
178
4

Find x.

Dec 18, 2018

#1
+700
0

Putting your question in LaTex, it is $$x^3 + (p+x)x^2 + 2px = 0$$.

Let's first factor this out. Using the distributive property, we have $$x^3 + px^2 + x^3 + 2px = 0 \Rightarrow 2x^3 + px^2 + 2px = 0$$.

We want to isolate p, so we subract 2x^3 from both sides to get $$px^2 + 2px = -2x^3$$. Factoring out p, we have $$p(x^2 + 2x) = -2x^3$$. Dividing x^2 + 2x from both sides, we have $$p = -\dfrac{2x^3}{x^2+2x} \Rightarrow \boxed{p = -\dfrac{2x^2}{x+2}}$$.

- PM

Dec 18, 2018
#2
+1

Solve for x:
2 p x + x^3 + x^2 (p + x) = 0

Expand out terms of the left hand side:
2 p x + p x^2 + 2 x^3 = 0

Factor x from the left hand side:
x (2 p + p x + 2 x^2) = 0

Split into two equations:
x = 0 or 2 p + p x + 2 x^2 = 0

Divide both sides by 2:
x = 0 or p + (p x)/2 + x^2 = 0

Subtract p from both sides:
x = 0 or (p x)/2 + x^2 = -p

x = 0 or p^2/16 + (p x)/2 + x^2 = p^2/16 - p
Write the left hand side as a square:
x = 0 or (p/4 + x)^2 = p^2/16 - p

Take the square root of both sides:
x = 0 or p/4 + x = sqrt(p^2/16 - p) or p/4 + x = -sqrt(p^2/16 - p)

Subtract p/4 from both sides:
x = 0 or x = sqrt((p^2)/16 - p) - p/4 or p/4 + x = -sqrt(p^2/16 - p)

Subtract p/4 from both sides:

x = 0     or     x = sqrt((p^2)/16 - p) - p/4    or     x = -p/4 - sqrt((p^2)/16 - p)

Dec 18, 2018
edited by Guest  Dec 18, 2018
#3
+103678
+2

x^3 + (p+x)x^2 +2px = 0

$$x^3 + (p+x)x^2 +2px = 0\\ x[x^2 + (p+x)x +2p]= 0\\ \text{x =0 is one solution}\\~\\ \text{If x is not 0 then}\\ x^2 + (p+x)x +2p=0\\ x^2 + x^2+px +2p=0\\ 2x^2 +px +2p=0\\ x=\frac{-p\pm\sqrt{p^2-4*2*2p}}{4}\\ x=\frac{-p\pm\sqrt{p^2-16p}}{4}\\~\\ so\\ x=\frac{-p\pm\sqrt{p^2-16p}}{4}\qquad or \qquad x=0$$

.
Dec 18, 2018