Solve the equation $x^3-3=2\sqrt{x+2}$.
I have tried to let $t=\sqrt{x+2}$ then we have $$\begin{cases} x^3-3&=2t \tag 1\\ t^2 &=x+2 \end{cases}$$ But I've stuck here...
Any help or advice on solving is much appreciated. Thanks!
Solve the equation $x^3-3=2\sqrt{x+2}$.
I have tried to let $t=\sqrt{x+2}$ then we have $$\begin{cases} x^3-3&=2t \tag 1\\ t^2 &=x+2 \end{cases}$$ But I've stuck here...
Any help or advice on solving is much appreciated. Thanks!
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$$x^3-3=2\sqrt{x+2}$$ $$x^6-6x^3+9=4x+8$$ $$x^6-6x^3-4x+1=0$$
At this point we get a sixth degree polynomial, and you can estimate the roots accordingly. Wolfram Alpha gives the positive solution as $x \approx 1.90874157798\dots$, but cannot find a closed form for the points (this can happen while solving any polynomial of degree $5$ of higher, by a proof by Galois)