Hello all, Need your help (Happy)

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

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- Apr 11th 2009, 04:28 PMlinda2005Solve this equation (involves radicals)
Hello all, Need your help (Happy)

Solve in $\displaystyle [-1,1]$ the equation : $\displaystyle \; (x+1)^2 = \sqrt{(1-x)^3} $ - Apr 11th 2009, 04:53 PMJhevon
- Apr 11th 2009, 05:01 PMlinda2005
Yes I do that , but $\displaystyle x(x^3 + 5x^2 + 6x + 7)=0$ it's so hard to resolve !

- Apr 11th 2009, 05:12 PMmr fantastic
- Apr 11th 2009, 05:12 PMProve It
It's actually

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

Clearly $\displaystyle x = 0$ is a solution.

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

Clearly it has one root at about $\displaystyle x = -4.65$, 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 PMNonCommAlg
for $\displaystyle x \geq -1$ we have: $\displaystyle x^3 + 5x^2 + 3x + 7 \geq x^3 + 3x + 7 \geq -1 - 3 + 7 = 3.$ so $\displaystyle x^3 + 5x

^2 + 3x + 7 \neq 0,$ for all $\displaystyle x \geq -1.$ - Apr 11th 2009, 05:24 PMmr fantastic
- Apr 11th 2009, 05:37 PMNonCommAlg