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Question Number 120769 by SOMEDAVONG last updated on 02/Nov/20
1).x^2 −5=(√(13+x)) ,Find x=? 2).x^3 +3x^2 −x−4=0 ,Find x=?
$$\left.\mathrm{1}\right).\mathrm{x}^{\mathrm{2}} −\mathrm{5}=\sqrt{\mathrm{13}+\mathrm{x}}\:\:,\mathrm{Find}\:\mathrm{x}=? \\ $$$$\left.\mathrm{2}\right).\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} −\mathrm{x}−\mathrm{4}=\mathrm{0}\:\:,\mathrm{Find}\:\mathrm{x}=? \\ $$
Answered by TANMAY PANACEA last updated on 02/Nov/20
2)f(x)=x^3 +3x^2 −x−4 f(0)<0 f(1)<0 f(2)>0 so one root lies between (1,2) 2>x>1 f(−1)<0 f(−2)>0 so one root lies between (−2,−1) f(−3)<0 so one root lies between (−3,−2)
$$\left.\mathrm{2}\right){f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −{x}−\mathrm{4} \\ $$$${f}\left(\mathrm{0}\right)<\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)<\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)>\mathrm{0} \\ $$$${so}\:{one}\:{root}\:{lies}\:{between}\:\left(\mathrm{1},\mathrm{2}\right)\:\:\:\:\mathrm{2}>{x}>\mathrm{1} \\ $$$${f}\left(−\mathrm{1}\right)<\mathrm{0} \\ $$$${f}\left(−\mathrm{2}\right)>\mathrm{0} \\ $$$${so}\:{one}\:{root}\:{lies}\:{between}\:\left(−\mathrm{2},−\mathrm{1}\right) \\ $$$${f}\left(−\mathrm{3}\right)<\mathrm{0} \\ $$$${so}\:{one}\:{root}\:{lies}\:{between}\:\left(−\mathrm{3},−\mathrm{2}\right) \\ $$
Commented by SOMEDAVONG last updated on 03/Nov/20
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Commented by TANMAY PANACEA last updated on 03/Nov/20
most welcome
$${most}\:{welcome} \\ $$
Answered by TANMAY PANACEA last updated on 02/Nov/20
1)x^2 −5=(√(13+x)) x=3 is a solution x can not be less than −13 f(x)=x^2 −5−(√(13+x)) f(0)<0 f(1)<0 f(2)<0 f(3)=0 f(4)>0 f(5)>0 f(−1)<0 f(−2)<0 f(−3)>0 so one root lies between (−3,−2)
$$\left.\mathrm{1}\right){x}^{\mathrm{2}} −\mathrm{5}=\sqrt{\mathrm{13}+{x}}\: \\ $$$${x}=\mathrm{3}\:{is}\:{a}\:{solution} \\ $$$${x}\:{can}\:{not}\:{be}\:{less}\:{than}\:−\mathrm{13} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{5}−\sqrt{\mathrm{13}+{x}}\: \\ $$$${f}\left(\mathrm{0}\right)<\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)<\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)<\mathrm{0} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{4}\right)>\mathrm{0} \\ $$$${f}\left(\mathrm{5}\right)>\mathrm{0} \\ $$$${f}\left(−\mathrm{1}\right)<\mathrm{0} \\ $$$${f}\left(−\mathrm{2}\right)<\mathrm{0} \\ $$$${f}\left(−\mathrm{3}\right)>\mathrm{0} \\ $$$${so}\:{one}\:{root}\:{lies}\:{between}\:\left(−\mathrm{3},−\mathrm{2}\right) \\ $$$$ \\ $$
Commented by SOMEDAVONG last updated on 03/Nov/20
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Commented by TANMAY PANACEA last updated on 03/Nov/20
most welcome
$${most}\:{welcome} \\ $$

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