Question Number 223995 by Rojarani last updated on 13/Aug/25
Answered by Frix last updated on 13/Aug/25
![Let y=px∧z=qx, solve for x^(−2) ⇒ x^(−2) = { ((7p^2 +9pq+3q^2 +13p+9q+7)),(((7p^2 +13pq+7q^2 +9p+9q+3)/4)),(((3p^2 +9pq+7q^2 +9p+13q+7)/9)) :} (1) = (2) ⇒ 21p^2 +23pq+5q^2 +43p+27q+25=0 (1) = (3) ⇒ 15p^2 +18pq+5q^2 +27p+17q+14=0 Subtract and solve for q ⇒ q=−((6p^2 +16p+11)/(5(p+2))) Insert above ⇒ (p+3)^2 (p^2 +p+1)=0 ⇒ p=−3 [ _(lead to false solutions)^(The complex values) ] ⇒ q=((17)/5) ⇒ x=±((5(√7))/(28))∧y=∓((15(√7))/(28))∧z=±((17(√7))/(28))](https://www.tinkutara.com/question/Q223996.png)
$$\mathrm{Let}\:{y}={px}\wedge{z}={qx},\:\mathrm{solve}\:\mathrm{for}\:{x}^{−\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}^{−\mathrm{2}} =\begin{cases}{\mathrm{7}{p}^{\mathrm{2}} +\mathrm{9}{pq}+\mathrm{3}{q}^{\mathrm{2}} +\mathrm{13}{p}+\mathrm{9}{q}+\mathrm{7}}\\{\left(\mathrm{7}{p}^{\mathrm{2}} +\mathrm{13}{pq}+\mathrm{7}{q}^{\mathrm{2}} +\mathrm{9}{p}+\mathrm{9}{q}+\mathrm{3}\right)/\mathrm{4}}\\{\left(\mathrm{3}{p}^{\mathrm{2}} +\mathrm{9}{pq}+\mathrm{7}{q}^{\mathrm{2}} +\mathrm{9}{p}+\mathrm{13}{q}+\mathrm{7}\right)/\mathrm{9}}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:=\:\left(\mathrm{2}\right)\:\Rightarrow\:\mathrm{21}{p}^{\mathrm{2}} +\mathrm{23}{pq}+\mathrm{5}{q}^{\mathrm{2}} +\mathrm{43}{p}+\mathrm{27}{q}+\mathrm{25}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:=\:\left(\mathrm{3}\right)\:\Rightarrow\:\mathrm{15}{p}^{\mathrm{2}} +\mathrm{18}{pq}+\mathrm{5}{q}^{\mathrm{2}} +\mathrm{27}{p}+\mathrm{17}{q}+\mathrm{14}=\mathrm{0} \\ $$$$\mathrm{Subtract}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:{q}\:\Rightarrow \\ $$$${q}=−\frac{\mathrm{6}{p}^{\mathrm{2}} +\mathrm{16}{p}+\mathrm{11}}{\mathrm{5}\left({p}+\mathrm{2}\right)} \\ $$$$\mathrm{Insert}\:\mathrm{above}\:\Rightarrow \\ $$$$\left({p}+\mathrm{3}\right)^{\mathrm{2}} \left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{p}=−\mathrm{3}\:\:\:\:\:\left[\:_{\mathrm{lead}\:\mathrm{to}\:\mathrm{false}\:\mathrm{solutions}} ^{\mathrm{The}\:\mathrm{complex}\:\mathrm{values}} \right]\:\:\:\:\:\Rightarrow\:{q}=\frac{\mathrm{17}}{\mathrm{5}} \\ $$$$\Rightarrow \\ $$$${x}=\pm\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{28}}\wedge{y}=\mp\frac{\mathrm{15}\sqrt{\mathrm{7}}}{\mathrm{28}}\wedge{z}=\pm\frac{\mathrm{17}\sqrt{\mathrm{7}}}{\mathrm{28}} \\ $$