Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 160806 by HongKing last updated on 06/Dec/21

Answered by MJS_new last updated on 06/Dec/21

A=0∨A=5    y=A−x  x^3 +(A−x)^3 =5x(A−x)  (3A+5)x^2 −A(3A+5)x+A^3 =0  x=(A/2)±((A(√(5−A)))/(2(√(3A+5))))  one solution if ((A(√(5−A)))/(2(√(3A+5))))=0 ⇒ A=0∨A=5

$${A}=\mathrm{0}\vee{A}=\mathrm{5} \\ $$$$ \\ $$$${y}={A}−{x} \\ $$$${x}^{\mathrm{3}} +\left({A}−{x}\right)^{\mathrm{3}} =\mathrm{5}{x}\left({A}−{x}\right) \\ $$$$\left(\mathrm{3}{A}+\mathrm{5}\right){x}^{\mathrm{2}} −{A}\left(\mathrm{3}{A}+\mathrm{5}\right){x}+{A}^{\mathrm{3}} =\mathrm{0} \\ $$$${x}=\frac{{A}}{\mathrm{2}}\pm\frac{{A}\sqrt{\mathrm{5}−{A}}}{\mathrm{2}\sqrt{\mathrm{3}{A}+\mathrm{5}}} \\ $$$$\mathrm{one}\:\mathrm{solution}\:\mathrm{if}\:\frac{{A}\sqrt{\mathrm{5}−{A}}}{\mathrm{2}\sqrt{\mathrm{3}{A}+\mathrm{5}}}=\mathrm{0}\:\Rightarrow\:{A}=\mathrm{0}\vee{A}=\mathrm{5} \\ $$

Commented by HongKing last updated on 06/Dec/21

cool thank you so much my dear Sir

$$\mathrm{cool}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com