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