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Question-66522




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

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