1. ## Solve this equation (involves radicals)

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

2. Originally Posted by linda2005

Solve in $[-1,1]$ the equation : $\; (x+1)^2 = \sqrt{(1-x)^3}$
Hint: square both sides and expand the brackets

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

4. Originally Posted by linda2005

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

5. 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

6. 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 $x \geq -1.$

7. 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 ....."

8. Originally Posted by linda2005

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.