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Question Number 171549 by Shrinava last updated on 17/Jun/22

Solve for real numbers: { ((2x^2 + 3y^2 + z^2 = 7)),((x^2 + y^2 + z^2 = (√2) z (x + y))) :}

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\mathrm{2x}^{\mathrm{2}} \:+\:\mathrm{3y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{7}}\\{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \:=\:\sqrt{\mathrm{2}}\:\mathrm{z}\:\left(\mathrm{x}\:+\:\mathrm{y}\right)}\end{cases} \\ $$

Answered by MJS_new last updated on 21/Jun/22

let y=px∧z=qx { (((3p^2 +q^2 +2)x^2 −7=0)),(((p^2 −(√2)pq+q^2 −(√2)q+1)x^2 =0)) :} ⇒ x=0∨(p^2 −(√2)pq+q^2 −(√2)q+1)=0 if x=0 the first equation is wrong and p^2 −(√2)pq+q^2 −(√2)q+1=0 ⇒ q=((√2)/2)(p+1±(p−1)i) ⇒ only real if p=1 ⇒ q=(√2) ⇒ (3+2+2)x^2 −7=0 ⇒ x=±1 ⇒ x=±1∧y=x∧z=(√2)x

$$\mathrm{let}\:{y}={px}\wedge{z}={qx} \\ $$$$\begin{cases}{\left(\mathrm{3}{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +\mathrm{2}\right){x}^{\mathrm{2}} −\mathrm{7}=\mathrm{0}}\\{\left({p}^{\mathrm{2}} −\sqrt{\mathrm{2}}{pq}+{q}^{\mathrm{2}} −\sqrt{\mathrm{2}}{q}+\mathrm{1}\right){x}^{\mathrm{2}} =\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\:{x}=\mathrm{0}\vee\left({p}^{\mathrm{2}} −\sqrt{\mathrm{2}}{pq}+{q}^{\mathrm{2}} −\sqrt{\mathrm{2}}{q}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{if}\:{x}=\mathrm{0}\:\mathrm{the}\:\mathrm{first}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{wrong}\:\mathrm{and} \\ $$$${p}^{\mathrm{2}} −\sqrt{\mathrm{2}}{pq}+{q}^{\mathrm{2}} −\sqrt{\mathrm{2}}{q}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:{q}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({p}+\mathrm{1}\pm\left({p}−\mathrm{1}\right)\mathrm{i}\right) \\ $$$$\Rightarrow\:\mathrm{only}\:\mathrm{real}\:\mathrm{if}\:{p}=\mathrm{1}\:\Rightarrow\:{q}=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{3}+\mathrm{2}+\mathrm{2}\right){x}^{\mathrm{2}} −\mathrm{7}=\mathrm{0}\:\Rightarrow\:{x}=\pm\mathrm{1} \\ $$$$\Rightarrow\:{x}=\pm\mathrm{1}\wedge{y}={x}\wedge{z}=\sqrt{\mathrm{2}}{x} \\ $$

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