Menu Close

Solve-for-x-x-2-c-c-x-

Tinku Tara 0 Comments
Pin




Question Number 193130 by Frix last updated on 04/Jun/23
Solve for x x^2 −c=(√(c−x))
$$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\mathrm{2}} −{c}=\sqrt{{c}−{x}} \\ $$
Commented by aba last updated on 04/Jun/23
t=c−x≥0 x^2 −(t−x)=(√t) ⇒x^2 −x−(t+(√t))=0 Δ=(2(√t)+1)^2 x_1 =((1−(2(√t)+1))/2)=−(√t) ∧ x_2 =((1+(2(√t)+1))/2)=1+(√t) ⧫x=−(√(c−x)) ⇒ x^2 +x−c=0 Δ=1+4c>0 x=±(1/2)((√(4c+1)) −1) ✓ ⧫x=1+(√(c−x)) ⇒ x−1=(√(c−x)) ⇒ x^2 −x+(1−c)=0 Δ=1+4(c−1)=4c−3>0 x_1 =((1−(√(4c−3)))/2) ∧ x=((1+(√(4c−3)))/2)✓ S={((1+(√(4c−3)))/2);±(1/2)((√(4c+1))−1)}
$$\mathrm{t}=\mathrm{c}−\mathrm{x}\geqslant\mathrm{0}\: \\ $$$$\mathrm{x}^{\mathrm{2}} −\left(\mathrm{t}−\mathrm{x}\right)=\sqrt{\mathrm{t}}\:\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\left(\mathrm{t}+\sqrt{\mathrm{t}}\right)=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{2}\sqrt{\mathrm{t}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{1}−\left(\mathrm{2}\sqrt{\mathrm{t}}+\mathrm{1}\right)}{\mathrm{2}}=−\sqrt{\mathrm{t}}\:\:\wedge\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{1}+\left(\mathrm{2}\sqrt{\mathrm{t}}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{1}+\sqrt{\mathrm{t}} \\ $$$$\blacklozenge\mathrm{x}=−\sqrt{\mathrm{c}−\mathrm{x}}\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{c}=\mathrm{0} \\ $$$$\Delta=\mathrm{1}+\mathrm{4c}>\mathrm{0} \\ $$$$\mathrm{x}=\pm\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{4c}+\mathrm{1}}\:−\mathrm{1}\right)\:\checkmark \\ $$$$\blacklozenge\mathrm{x}=\mathrm{1}+\sqrt{\mathrm{c}−\mathrm{x}}\:\:\Rightarrow\:\mathrm{x}−\mathrm{1}=\sqrt{\mathrm{c}−\mathrm{x}}\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\left(\mathrm{1}−\mathrm{c}\right)=\mathrm{0} \\ $$$$\Delta=\mathrm{1}+\mathrm{4}\left(\mathrm{c}−\mathrm{1}\right)=\mathrm{4c}−\mathrm{3}>\mathrm{0} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{1}−\sqrt{\mathrm{4c}−\mathrm{3}}}{\mathrm{2}}\:\wedge\:\mathrm{x}=\frac{\mathrm{1}+\sqrt{\mathrm{4c}−\mathrm{3}}}{\mathrm{2}}\checkmark \\ $$$$\mathrm{S}=\left\{\frac{\mathrm{1}+\sqrt{\mathrm{4c}−\mathrm{3}}}{\mathrm{2}};\pm\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{4c}+\mathrm{1}}−\mathrm{1}\right)\right\} \\ $$
Answered by Mastermind last updated on 04/Jun/23
Solution take square of both sides, we have ⇒(x^2 −c)^2 =c−x ⇒x^4 −2x^2 c+c^2 =c−x ⇒c^2 −c−2x^2 c+x^4 −x=0 ⇒c^2 −(1+2x^2 )+x^4 −x=0 ∙∙∙∙∙∗ ⇒solve equation ∗ by using Almighty formular (Quadratic formula) c= ((−b±(√(b^2 −4ac)))/(2a)) where b=−(1+2x^2 ), a=1 and c=x^4 −4 therefore, c=(((1+2x^2 )±(√((1+4x^2 +4x^4 )−4(x^4 −x))))/(2×1)) c=(((1+2x^2 ))/2) ± ((√(1+5x^2 ))/2)
$$\mathrm{Solution} \\ $$$$\mathrm{take}\:\mathrm{square}\:\mathrm{of}\:\mathrm{both}\:\mathrm{sides},\:\mathrm{we}\:\mathrm{have} \\ $$$$\Rightarrow\left(\mathrm{x}^{\mathrm{2}} −\mathrm{c}\right)^{\mathrm{2}} =\mathrm{c}−\mathrm{x} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{2}} \mathrm{c}+\mathrm{c}^{\mathrm{2}} =\mathrm{c}−\mathrm{x} \\ $$$$\Rightarrow\mathrm{c}^{\mathrm{2}} −\mathrm{c}−\mathrm{2x}^{\mathrm{2}} \mathrm{c}+\mathrm{x}^{\mathrm{4}} −\mathrm{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{c}^{\mathrm{2}} −\left(\mathrm{1}+\mathrm{2x}^{\mathrm{2}} \right)+\mathrm{x}^{\mathrm{4}} −\mathrm{x}=\mathrm{0}\:\centerdot\centerdot\centerdot\centerdot\centerdot\ast \\ $$$$\Rightarrow\mathrm{solve}\:\mathrm{equation}\:\ast\:\mathrm{by}\:\mathrm{using}\:\mathrm{Almighty} \\ $$$$\mathrm{formular}\:\left(\mathrm{Quadratic}\:\mathrm{formula}\right) \\ $$$$\mathrm{c}=\:\frac{−\mathrm{b}\pm\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}}{\mathrm{2a}} \\ $$$$\mathrm{where}\:\mathrm{b}=−\left(\mathrm{1}+\mathrm{2x}^{\mathrm{2}} \right),\:\mathrm{a}=\mathrm{1}\:\mathrm{and}\:\mathrm{c}=\mathrm{x}^{\mathrm{4}} −\mathrm{4} \\ $$$$ \\ $$$$\mathrm{therefore},\: \\ $$$$\mathrm{c}=\frac{\left(\mathrm{1}+\mathrm{2x}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}^{\mathrm{4}} \right)−\mathrm{4}\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}\right)}}{\mathrm{2}×\mathrm{1}} \\ $$$$\mathrm{c}=\frac{\left(\mathrm{1}+\mathrm{2x}^{\mathrm{2}} \right)}{\mathrm{2}}\:\pm\:\frac{\sqrt{\mathrm{1}+\mathrm{5x}^{\mathrm{2}} }}{\mathrm{2}} \\ $$
Answered by aba last updated on 04/Jun/23
x=±(1/2)((√(4c+1))−1) ∨ x=(1/2)((√(4x−3))+1)
$$\mathrm{x}=\pm\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{4c}+\mathrm{1}}−\mathrm{1}\right)\:\vee\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{4x}−\mathrm{3}}+\mathrm{1}\right) \\ $$
Answered by Frix last updated on 04/Jun/23
Finding one solution using f(x)=f^(−1) (x) y=(√(c−x)) ⇔ x=c−x^2 =−(x^2 −c) ⇒ −x=(√(c−x)) ⇔ x^2 =c−x∧x<0 ⇒ x_1 =−((1+(√(4c+1)))/2) [x_2 =−((1−(√(4c+1)))/2)≥0] x^2 −c=(√(c−x)) parabola = positive half parabola ⇒ 2 solutions Solving for c is easy (x^2 −c)^2 =c−x c^2 −(2x^2 +1)c+x(x^3 +1)=0 c=((2x^2 +1)/2)±((2x−1)/2) c=x^2 −x+1∨c=x^2 +x Now solve these for x x=((1±(√(4c−3)))/2)∨x=−((1±(√(4c+1)))/2) But 2 solutions are false x_1 =−((1+(√(4c+1)))/2) [from above] x_2 =((1+(√(4c−3)))/2) [because it must be >0 due to symmetry of x^2 −c]
$$\mathrm{Finding}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{using}\:{f}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right) \\ $$$${y}=\sqrt{{c}−{x}}\:\Leftrightarrow\:{x}={c}−{x}^{\mathrm{2}} =−\left({x}^{\mathrm{2}} −{c}\right) \\ $$$$\Rightarrow \\ $$$$−{x}=\sqrt{{c}−{x}}\:\Leftrightarrow\:{x}^{\mathrm{2}} ={c}−{x}\wedge{x}<\mathrm{0}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} =−\frac{\mathrm{1}+\sqrt{\mathrm{4}{c}+\mathrm{1}}}{\mathrm{2}}\:\:\:\:\:\left[{x}_{\mathrm{2}} =−\frac{\mathrm{1}−\sqrt{\mathrm{4}{c}+\mathrm{1}}}{\mathrm{2}}\geqslant\mathrm{0}\right] \\ $$$$ \\ $$$${x}^{\mathrm{2}} −{c}=\sqrt{{c}−{x}} \\ $$$$\mathrm{parabola}\:=\:\mathrm{positive}\:\mathrm{half}\:\mathrm{parabola}\:\Rightarrow\:\mathrm{2}\:\mathrm{solutions} \\ $$$$\mathrm{Solving}\:\mathrm{for}\:{c}\:\mathrm{is}\:\mathrm{easy} \\ $$$$\left({x}^{\mathrm{2}} −{c}\right)^{\mathrm{2}} ={c}−{x} \\ $$$${c}^{\mathrm{2}} −\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right){c}+{x}\left({x}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${c}=\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{2}} \\ $$$${c}={x}^{\mathrm{2}} −{x}+\mathrm{1}\vee{c}={x}^{\mathrm{2}} +{x} \\ $$$$\mathrm{Now}\:\mathrm{solve}\:\mathrm{these}\:\mathrm{for}\:{x} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{4}{c}−\mathrm{3}}}{\mathrm{2}}\vee{x}=−\frac{\mathrm{1}\pm\sqrt{\mathrm{4}{c}+\mathrm{1}}}{\mathrm{2}} \\ $$$$\mathrm{But}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{false} \\ $$$${x}_{\mathrm{1}} =−\frac{\mathrm{1}+\sqrt{\mathrm{4}{c}+\mathrm{1}}}{\mathrm{2}}\:\left[\mathrm{from}\:\mathrm{above}\right] \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}+\sqrt{\mathrm{4}{c}−\mathrm{3}}}{\mathrm{2}}\:\left[\mathrm{because}\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:>\mathrm{0}\:\mathrm{due}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{to}\:\mathrm{symmetry}\:\mathrm{of}\:{x}^{\mathrm{2}} −{c}\right] \\ $$
Commented by aba last updated on 04/Jun/23
����

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *