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Question Number 55312 by Joel578 last updated on 21/Feb/19

$$\mathrm{Consider}\mathrm{the}\mathrm{system}$$$$\{\begin{array}{l}{x}^2+y^2=z\\ 2x+2y+z=k\end{array}$$$$\mathrm{The}\mathrm{value}\mathrm{of}xy+zk\mathrm{for}\mathrm{which}\mathrm{the}\mathrm{system}$$$$\mathrm{has}\mathrm{a}\mathrm{unique}\mathrm{solution}\mathrm{is}...$$

Answered by mr W last updated on 21/Feb/19

$$2^{2}z+2^{2}z={(z-k)}^2$$$$z^2-2(k+4)z+k^2=0$$$$4{(k+4)}^2-4k^2=0$$$$\Rightarrow k=-2$$$$\Rightarrow z=2$$$$x^2+y^2=2$$$$x+y=-2$$$$\Rightarrow x=-1$$$$\Rightarrow y=-1$$$$\Rightarrow xy+kz=1-4=-3$$

Commented by Joel578 last updated on 22/Feb/19

$$thankyouverymuch$$

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