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Question Number 8939 by arinto27 last updated on 07/Nov/16
1) diket lingkaran c dg pers x^2 +y^2 −6x+2y+2=0  diket pula titik d (a,−3). agar d brd di dlm lingkaran  nilai a yg memenuhi adalah....
$$\left.\mathrm{1}\right)\:\mathrm{diket}\:\mathrm{lingkaran}\:\mathrm{c}\:\mathrm{dg}\:\mathrm{pers}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{6x}+\mathrm{2y}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{diket}\:\mathrm{pula}\:\mathrm{titik}\:\mathrm{d}\:\left(\mathrm{a},−\mathrm{3}\right).\:\mathrm{agar}\:\mathrm{d}\:\mathrm{brd}\:\mathrm{di}\:\mathrm{dlm}\:\mathrm{lingkaran} \\ $$$$\mathrm{nilai}\:\mathrm{a}\:\mathrm{yg}\:\mathrm{memenuhi}\:\mathrm{adalah}…. \\ $$
Answered by sandy_suhendra last updated on 09/Nov/16
a^2 +(−3)^2 −6a+2(−3)+2 < 0  a^2 +9−6a−6+2 < 0  a^2 −6a+5 < 0  (a−5)(a−1) < 0  Gunakan grs bilangan dan tes tanda  akan diperoleh : 1<a<5
$$\mathrm{a}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{6a}+\mathrm{2}\left(−\mathrm{3}\right)+\mathrm{2}\:<\:\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{9}−\mathrm{6a}−\mathrm{6}+\mathrm{2}\:<\:\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{6a}+\mathrm{5}\:<\:\mathrm{0} \\ $$$$\left(\mathrm{a}−\mathrm{5}\right)\left(\mathrm{a}−\mathrm{1}\right)\:<\:\mathrm{0} \\ $$$$\mathrm{Gunakan}\:\mathrm{grs}\:\mathrm{bilangan}\:\mathrm{dan}\:\mathrm{tes}\:\mathrm{tanda} \\ $$$$\mathrm{akan}\:\mathrm{diperoleh}\::\:\mathrm{1}<\mathrm{a}<\mathrm{5} \\ $$

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