# Solve this equation (involves radicals)

• Apr 11th 2009, 04:28 PM
linda2005
Solve this equation (involves radicals)
Hello all, Need your help (Happy)

Solve in $[-1,1]$ the equation : $\; (x+1)^2 = \sqrt{(1-x)^3}$
• Apr 11th 2009, 04:53 PM
Jhevon
Quote:

Originally Posted by linda2005
Hello all, Need your help (Happy)

Solve in $[-1,1]$ the equation : $\; (x+1)^2 = \sqrt{(1-x)^3}$

Hint: square both sides and expand the brackets
• Apr 11th 2009, 05:01 PM
linda2005
Yes I do that , but $x(x^3 + 5x^2 + 6x + 7)=0$ it's so hard to resolve !
• Apr 11th 2009, 05:12 PM
mr fantastic
Quote:

Originally Posted by linda2005
Hello all, Need your help

Solve in ${\color{red}[-1,1]}$ the equation : $\; (x+1)^2 = \sqrt{(1-x)^3}$

Quote:

Originally Posted by linda2005
Yes I do that , but $x(x^3 + 5x^2 + {\color{red}3} x + 7)=0$ it's so hard to resolve ! Mr F says: Note the red.

The key phrase in the question is "Solve in [-1, 1] ....". It is clear that the (only) solution to the cubic lies outside this domain ....

So the only solution is x = 0.
• Apr 11th 2009, 05:12 PM
Prove It
Quote:

Originally Posted by linda2005
Yes I do that , but $x(x^3 + 5x^2 + 6x + 7)=0$ it's so hard to resolve !

It's actually

$x(x^3 + 5x^2 + 3x + 7) = 0$.

Clearly $x = 0$ is a solution.

Try graphing the equation $x^3 + 5x^2 + 3x + 7$.

Clearly it has one root at about $x = -4.65$, but I don't believe there's a way to solve it exactly (I may be wrong).

You might want to use a Heuristic method like the Bisection Method or Newton's Method to increase the accuracy.

Edit: And as Mr F says, it doesn't lie in the domain anyway... this post is for future reference :P
• Apr 11th 2009, 05:22 PM
NonCommAlg
for $x \geq -1$ we have: $x^3 + 5x^2 + 3x + 7 \geq x^3 + 3x + 7 \geq -1 - 3 + 7 = 3.$ so $x^3 + 5x
^2 + 3x + 7 \neq 0,$
for all $x \geq -1.$
• Apr 11th 2009, 05:24 PM
mr fantastic
Quote:

Originally Posted by NonCommAlg
for $x \geq -1$ we have: $x^3 + 5x^2 + 3x + 7 \geq x^3 + 3x + 7 \geq -1 - 3 + 7 = 3.$ so $x^3 + 5x
^2 + 3x + 7 \neq 0,$
for all $x \geq -1.$

Like I said ... "It is clear ....." (Rofl)
• Apr 11th 2009, 05:37 PM
NonCommAlg
Quote:

Originally Posted by linda2005

Hello all, Need your help (Happy)

Solve in $[-1,1]$ the equation : $\; (x+1)^2 = \sqrt{(1-x)^3}$

was $x \geq -1$ a part of problem or ...? because the only restriction that we need here is $x \leq 1$ to make sure that the right hand side of the equation is defined.