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Question Number 73399 by mathocean1 last updated on 11/Nov/19

$$\{\begin{array}{l}{x}^2+y^2=65\\ (x-1)(y-1)=17\end{array}$$$$$$$$pleasehelpmetosolveit...$$

Commented by abdomathmax last updated on 11/Nov/19

$$\Rightarrow \{\begin{array}{l}{(x+y)}^2-2xy=65\\ xy-(x+y)+1=17\Rightarrow \{\begin{array}{l}{s}^2-2p=65\\ p-s=16\Rightarrow \end{array}\end{array}$$$$\{\begin{array}{l}{s}^2-2p=65\\ 2p=2s+32\Rightarrow \{\begin{array}{l}{s}^2-(2s+32)=65\\ p=s+16\Rightarrow \end{array}\end{array}$$$$\{\begin{array}{l}{s}^2-2s-32-65=0\\ p=s+16\Rightarrow \{\begin{array}{l}{s}^2-2s-97=0\\ p=s+16\end{array}\end{array}$$$$letsolve{s}^2-2s-97=0$$$${\mathrm{\Delta }}^{{}^{\prime }}=98\Rightarrow s_1=1+\sqrt{98}and{s}_2=1-\sqrt{98}$$$$s_1=1+\sqrt{98}\Rightarrow p_1=17+\sqrt{98}$$$$s_2=1-\sqrt{98}\Rightarrow p_2=17-\sqrt{98}$$$$wehavex+y=sandxy=p\Rightarrow xandyaresolution?of$$$$theequation{X}^2-sX+p=0$$$$s=s_1\Rightarrow X^2-(1+\sqrt{98})X+17+\sqrt{98}=0$$$$\mathrm{\Delta }={(1+\sqrt{98})}^2-4(17+\sqrt{98})=99+2\sqrt{98}-68-4\sqrt{98}$$$$=31-2\sqrt{98}$$$$x=\frac{1+\sqrt{98}+\sqrt{31-2\sqrt{98}}}{2}$$$$y=\frac{1+\sqrt{98}-\sqrt{31-2\sqrt{98}}}{2}$$$$s=s_2\Rightarrow X^2-(1-\sqrt{98})X+(17-\sqrt{98})=0$$$$andwefollowthesameway...$$

Answered by $@ty@m123 last updated on 12/Nov/19

$$(x-1)(y-1)=17$$$$xy-x-y+1=17$$$$2xy-2(x+y)=32....\left(1\right)$$$$x^2+y^2=65...\left(2\right)$$$$Adding$$$$x^2+y^2+2xy-2(x+y)=97$$$$k^2-2k-97=0wherek=x+y$$$$k=\frac{2\pm \sqrt{4+388}}{2}$$$$k=\frac{2\pm \sqrt{392}}{2}$$$$k=\frac{2\pm 19.8}{2}$$$$k=\frac{21.8}{2},\frac{-17.8}{2}$$$$k=10.9,-8.9....\left(3\right)$$$$Subtracting\left(1\right)from\left(2\right),$$$$x^2+y^2-2xy+2(x+y)=33$$$${(x-y)}^2+2(x+y)=33.....\left(4\right)$$$$Case\left(i\right)k=10.9orx+y=10.9...\left(5\right)$$$${(x-y)}^2+21.8=33$$$${(x-y)}^2=11.2\Rightarrow x-y=\sqrt{11.2}....\left(6\right)$$$$Nowsolve\left(5\right)and\left(6\right)$$$$$$$$Case\left(ii\right)k=-8.9orx+y=-8.9...\left(7\right)$$$${(x-y)}^2-17.8=15$$$${(x-y)}^2=17.8+15$$$${(x-y)}^2=32.8$$$$x-y=\sqrt{32.8}...\left(8\right)$$$$Nowsolve\left(7\right)and\left(8\right)$$

Answered by ajfour last updated on 11/Nov/19

$$x-1=p,y-1=q$$$$pq=17$$$${(p+1)}^2+{(q+1)}^2=65$$$${(p+q)}^2-2pq+2(p+q)=63$$$${(p+q)}^2+2(p+q)-97=0$$$$p+q=-1\pm 7\sqrt{2}=c$$$$p,qarerootsof$$$$z^2-cz+17=0$$$$z=\frac{c}{2}\pm \sqrt{\frac{c^2}{4}-17}$$$$x=p+1=1+\frac{c}{2}\pm \sqrt{\frac{c^2}{4}-17}$$$$y=q+1=1+\frac{c}{2}\mp \sqrt{\frac{c^2}{4}-17}$$$$c=-1\pm 7\sqrt{2}.$$

Commented by mind is power last updated on 11/Nov/19

$$hellosirhaveyoupost12houragosomthingsboutapolynomialofdeg3\int ?$$$$iseeitboutiwasstworckcanyoureploaditplease?$$$$$$

Commented by ajfour last updated on 12/Nov/19

$$it{wasn}^{\prime }tastandardquestion,$$$$ishalluploadanotherbetterone$$$$andrequestyoutosolve..$$

Answered by MJS last updated on 11/Nov/19

$$x=p-q$$$$y=p+q$$$$\left(1\right)2p^2+2q^2=65\Rightarrow q^2=\frac{65}{2}-p^2$$$$\left(2\right)p^2-2p-q^2+1=17\Rightarrow q^2=p^2-2p-16$$$$$$$$\frac{65}{2}-p^2=p^2-2p-16$$$$\Rightarrow p_1=\frac{1-7\sqrt{2}}{2};p_2=\frac{1+7\sqrt{2}}{2}$$$$\Rightarrow q_1=\pm \frac{\sqrt{31+14\sqrt{2}}}{2};q_2=\pm \frac{\sqrt{31-14\sqrt{2}}}{2}$$$$\Rightarrow$$$$x_1=\frac{1-7\sqrt{2}}{2}\pm \frac{\sqrt{31+14\sqrt{2}}}{2};y_1=\frac{1-7\sqrt{2}}{2}\mp \frac{\sqrt{31+14\sqrt{2}}}{2}$$$$x_2=\frac{1+7\sqrt{2}}{2}\pm \frac{\sqrt{31-14\sqrt{2}}}{2};y_2=\frac{1+7\sqrt{2}}{2}\mp \frac{\sqrt{31-14\sqrt{2}}}{2}$$

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