Question Number 66522 by mr W last updated on 16/Aug/19
Answered by MJS last updated on 16/Aug/19
$${q}^{\mathrm{2}} +{x}^{\mathrm{2}} =\mathrm{16} \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} =\mathrm{25} \\ $$$$\left({x}−{q}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} =\mathrm{9} \\ $$$$ \\ $$$$\mathrm{solving}\:\mathrm{leads}\:\mathrm{to} \\ $$$${p}=\frac{\mathrm{3}\sqrt{\mathrm{17}}}{\mathrm{17}} \\ $$$${q}=\frac{\mathrm{4}\sqrt{\mathrm{17}}}{\mathrm{17}} \\ $$$${x}=\frac{\mathrm{16}\sqrt{\mathrm{17}}}{\mathrm{17}} \\ $$
Commented by MJS last updated on 16/Aug/19
$$…\mathrm{leading}\:\mathrm{to}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{degree}\:\mathrm{for}\:{v}, \\ $$$$\mathrm{or}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{degree}\:\mathrm{for}\:{v}^{\mathrm{2}} \:\left({v}=\mathrm{any}\:\mathrm{of}\:{p},\:{q}\:\mathrm{or}\:{x}\right)\:\mathrm{if}\:\mathrm{you} \\ $$$$\mathrm{do}\:\mathrm{it}\:\mathrm{right}… \\ $$
Answered by MJS last updated on 16/Aug/19
$$\mathrm{4sin}\:\alpha\:={q} \\ $$$$\mathrm{4cos}\:\alpha\:={x} \\ $$$$\mathrm{3sin}\:\alpha\:={p} \\ $$$$\mathrm{3cos}\:\alpha\:={x}−{q} \\ $$$$\Rightarrow \\ $$$$\frac{{q}}{\mathrm{4}}=\frac{{p}}{\mathrm{3}} \\ $$$$\frac{{x}}{\mathrm{4}}=\frac{{x}−{q}}{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$${p}=\frac{\mathrm{3}{x}}{\mathrm{16}} \\ $$$${q}=\frac{{x}}{\mathrm{4}} \\ $$$$ \\ $$$${q}^{\mathrm{2}} +{x}^{\mathrm{2}} =\mathrm{16} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{16}}+{x}^{\mathrm{2}} =\mathrm{16} \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{16}\sqrt{\mathrm{17}}}{\mathrm{17}} \\ $$
Commented by mr W last updated on 16/Aug/19
$${thanks}\:{sir}! \\ $$
Commented by MJS last updated on 16/Aug/19
$$\mathrm{easier}… \\ $$
Answered by mr W last updated on 16/Aug/19
$$\sqrt{\mathrm{4}^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}}{x}={x} \\ $$$$\sqrt{\mathrm{4}^{\mathrm{2}} −{x}^{\mathrm{2}} }=\frac{{x}}{\mathrm{4}} \\ $$$$\mathrm{4}^{\mathrm{2}} −{x}^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\mathrm{4}^{\mathrm{2}} =\frac{\mathrm{17}{x}^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{16}}{\:\sqrt{\mathrm{17}}} \\ $$