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Question Number 193130 by Frix last updated on 04/Jun/23

$$\mathrm{Solve}\mathrm{for}x$$$$x^2-c=\sqrt{c-x}$$

Commented by aba last updated on 04/Jun/23

$$\mathrm{t}=\mathrm{c}-\mathrm{x}⩾0$$$${\mathrm{x}}^2-(\mathrm{t}-\mathrm{x})=\sqrt{\mathrm{t}}\Rightarrow {\mathrm{x}}^2-\mathrm{x}-(\mathrm{t}+\sqrt{\mathrm{t}})=0$$$$\mathrm{\Delta }={(2\sqrt{\mathrm{t}}+1)}^2$$$${\mathrm{x}}_1=\frac{1-(2\sqrt{\mathrm{t}}+1)}{2}=-\sqrt{\mathrm{t}}\wedge {\mathrm{x}}_2=\frac{1+(2\sqrt{\mathrm{t}}+1)}{2}=1+\sqrt{\mathrm{t}}$$$$⧫\mathrm{x}=-\sqrt{\mathrm{c}-\mathrm{x}}\Rightarrow {\mathrm{x}}^2+\mathrm{x}-\mathrm{c}=0$$$$\mathrm{\Delta }=1+4\mathrm{c}>0$$$$\mathrm{x}=\pm \frac{1}{2}(\sqrt{4\mathrm{c}+1}-1)✓$$$$⧫\mathrm{x}=1+\sqrt{\mathrm{c}-\mathrm{x}}\Rightarrow \mathrm{x}-1=\sqrt{\mathrm{c}-\mathrm{x}}\Rightarrow {\mathrm{x}}^2-\mathrm{x}+(1-\mathrm{c})=0$$$$\mathrm{\Delta }=1+4(\mathrm{c}-1)=4\mathrm{c}-3>0$$$${\mathrm{x}}_1=\frac{1-\sqrt{4\mathrm{c}-3}}{2}\wedge \mathrm{x}=\frac{1+\sqrt{4\mathrm{c}-3}}{2}✓$$$$\mathrm{S}=\{\frac{1+\sqrt{4\mathrm{c}-3}}{2};\pm \frac{1}{2}(\sqrt{4\mathrm{c}+1}-1)\}$$

Answered by Mastermind last updated on 04/Jun/23

$$\mathrm{Solution}$$$$\mathrm{take}\mathrm{square}\mathrm{of}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{have}$$$$\Rightarrow {({\mathrm{x}}^2-\mathrm{c})}^2=\mathrm{c}-\mathrm{x}$$$$\Rightarrow {\mathrm{x}}^4-{2\mathrm{x}}^2\mathrm{c}+{\mathrm{c}}^2=\mathrm{c}-\mathrm{x}$$$$\Rightarrow {\mathrm{c}}^2-\mathrm{c}-{2\mathrm{x}}^2\mathrm{c}+{\mathrm{x}}^4-\mathrm{x}=0$$$$\Rightarrow {\mathrm{c}}^2-(1+{2\mathrm{x}}^2)+{\mathrm{x}}^4-\mathrm{x}=0\cdot \cdot \cdot \cdot \cdot \ast$$$$\Rightarrow \mathrm{solve}\mathrm{equation}\ast \mathrm{by}\mathrm{using}\mathrm{Almighty}$$$$\mathrm{formular}\left(\mathrm{Quadratic}\mathrm{formula}\right)$$$$\mathrm{c}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$$$$\mathrm{where}\mathrm{b}=-(1+{2\mathrm{x}}^2),\mathrm{a}=1\mathrm{and}\mathrm{c}={\mathrm{x}}^4-4$$$$$$$$\mathrm{therefore},$$$$\mathrm{c}=\frac{(1+{2\mathrm{x}}^2)\pm \sqrt{(1+{4\mathrm{x}}^2+{4\mathrm{x}}^4)-4({\mathrm{x}}^4-\mathrm{x})}}{2\times 1}$$$$\mathrm{c}=\frac{(1+{2\mathrm{x}}^2)}{2}\pm \frac{\sqrt{1+{5\mathrm{x}}^2}}{2}$$

Answered by aba last updated on 04/Jun/23

$$\mathrm{x}=\pm \frac{1}{2}(\sqrt{4\mathrm{c}+1}-1)\vee \mathrm{x}=\frac{1}{2}(\sqrt{4\mathrm{x}-3}+1)$$

Answered by Frix last updated on 04/Jun/23

$$\mathrm{Finding}\mathrm{one}\mathrm{solution}\mathrm{using}f\left(x\right)=f^{-1}\left(x\right)$$$$y=\sqrt{c-x}\iff x=c-x^2=-(x^2-c)$$$$\Rightarrow$$$$-x=\sqrt{c-x}\iff x^2=c-x\wedge x<0\Rightarrow$$$$x_1=-\frac{1+\sqrt{4c+1}}{2}[x_2=-\frac{1-\sqrt{4c+1}}{2}⩾0]$$$$$$$$x^2-c=\sqrt{c-x}$$$$\mathrm{parabola}=\mathrm{positive}\mathrm{half}\mathrm{parabola}\Rightarrow 2\mathrm{solutions}$$$$\mathrm{Solving}\mathrm{for}c\mathrm{is}\mathrm{easy}$$$${(x^2-c)}^2=c-x$$$$c^2-(2x^2+1)c+x(x^3+1)=0$$$$c=\frac{2x^2+1}{2}\pm \frac{2x-1}{2}$$$$c=x^2-x+1\vee c=x^2+x$$$$\mathrm{Now}\mathrm{solve}\mathrm{these}\mathrm{for}x$$$$x=\frac{1\pm \sqrt{4c-3}}{2}\vee x=-\frac{1\pm \sqrt{4c+1}}{2}$$$$\mathrm{But}2\mathrm{solutions}\mathrm{are}\mathrm{false}$$$$x_1=-\frac{1+\sqrt{4c+1}}{2}\left[\mathrm{from}\mathrm{above}\right]$$$$x_2=\frac{1+\sqrt{4c-3}}{2}[\mathrm{because}\mathrm{it}\mathrm{must}\mathrm{be}>0\mathrm{due}$$$$\mathrm{to}\mathrm{symmetry}\mathrm{of}{x}^2-c]$$

Commented by aba last updated on 04/Jun/23

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