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Question Number 56878 by ajfour last updated on 25/Mar/19

$$\mathrm{f}\left(\mathrm{x}\right)=\mid \mid \mathrm{x}\mid -4\mid$$$$\mathrm{g}\left(\mathrm{x}\right)=\sqrt{36-{\mathrm{x}}^2}-3$$$$\mathrm{Find}\mathrm{points}\mathrm{of}\mathrm{intersection}.$$

Answered by mr W last updated on 25/Mar/19

$$lett=\mid x\mid ,0⩽t⩽6$$$$ift⩽4:$$$$\sqrt{36-t^2}-3=4-t$$$$\sqrt{36-t^2}=7-t$$$$36-t^2=49-14t+t^2$$$$2t^2-14t+13=0$$$$t=\frac{7-\sqrt{23}}{2}<4\Rightarrow ok$$$$\Rightarrow x=\pm \frac{7-\sqrt{23}}{2}$$$$$$$$if4⩽t⩽6:$$$$\sqrt{36-t^2}-3=t-4$$$$\sqrt{36-t^2}=t-1$$$$36-t^2=t^2-2t+1$$$$2t^2-2t-35=0$$$$t=\frac{1+\sqrt{71}}{2}>4\Rightarrow ok$$$$\Rightarrow x=\pm \frac{1+\sqrt{71}}{2}$$

Commented by ajfour last updated on 26/Mar/19

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}.$$

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