Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 55198 by ajfour last updated on 19/Feb/19

x^5 −4x^4 +6x^3 +8x−32=0  Find at least one root.

$${x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}−\mathrm{32}=\mathrm{0} \\ $$$${Find}\:{at}\:{least}\:{one}\:{root}. \\ $$

Commented by arvinddayama00@gmail.com last updated on 19/Feb/19

all velue of x=?

$${all}\:{velue}\:{of}\:{x}=? \\ $$

Answered by rahul 19 last updated on 19/Feb/19

x=2!

$${x}=\mathrm{2}! \\ $$

Commented by rahul 19 last updated on 19/Feb/19

simply by hit & trial . ����

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19

x^5 +6x^3 +8x=4x^4 +32  right hand side x^(even power)  so right hand  side is always +ve  now left hand side should be +ve  so x can not be less than zero  so x≮0  now    (x^5 +6x^3 +8x)−(4x^4 +32)=0  f(x)=(x^5 +6x^3 +8x)−(4x^4 +32)  f(x)<0 at x=0  f(x)<0 at x=1  so no root in between(0,1)    now checking  between 1 and 2  f(2)=32+48+16−64−32=0  so x=2 is a root..  f(1)=1+6+8−4−32=−21  now gradually we increase the value of x  such as 1.1,1.2,...and observe change of sign  f(x)=y=(x^5 +6x^3 +8x)−(4x^4 +32)  (dy/dx)=5x^4 +18x^2 +8−16x^3   ((△y)/(△x))≈(dy/dx)  △y=(dy/dx)×△x=(5+18+8−16)×0.1=1.5  so f(1.1)=f(1)+△y=−21+1.5=−19.5  f(2.1)=f(2)+△y=0+1.5=1.5>0  f(3)=3^5 +6×3^3 +8×3−4×3^4 −32  =3^4 (3−4)+6×27−8  =−81+162−8>0  so no root between 2.1 and 3  ...  f(4)=4^5 +6×4^3 +32−4×4^4 −32>0  so no root between 3 and 4  now...  f(x)=x^5 +6x^3 +8x−(4x^4 +32)  f(x)=g(x)−h(x)  (df/dx)=(dg/dx)−(dh/dx)=(5x^4 +18x^2 +8)−(16x^3 )  =x^3 (5x−16)+18x^2 +8..  now look when x>3.2    (df/dx)>0  so there will be no sign change when x>3.2  so this eqn has only one root that is  x=2...pls pay your kind attention...

$${x}^{\mathrm{5}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}=\mathrm{4}{x}^{\mathrm{4}} +\mathrm{32} \\ $$$${right}\:{hand}\:{side}\:{x}^{{even}\:{power}} \:{so}\:{right}\:{hand} \\ $$$${side}\:{is}\:{always}\:+{ve} \\ $$$${now}\:{left}\:{hand}\:{side}\:{should}\:{be}\:+{ve} \\ $$$${so}\:{x}\:{can}\:{not}\:{be}\:{less}\:{than}\:{zero} \\ $$$${so}\:{x}\nless\mathrm{0} \\ $$$${now}\:\: \\ $$$$\left({x}^{\mathrm{5}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}\right)−\left(\mathrm{4}{x}^{\mathrm{4}} +\mathrm{32}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{5}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}\right)−\left(\mathrm{4}{x}^{\mathrm{4}} +\mathrm{32}\right) \\ $$$${f}\left({x}\right)<\mathrm{0}\:{at}\:{x}=\mathrm{0} \\ $$$${f}\left({x}\right)<\mathrm{0}\:{at}\:{x}=\mathrm{1} \\ $$$${so}\:{no}\:{root}\:{in}\:{between}\left(\mathrm{0},\mathrm{1}\right) \\ $$$$ \\ $$$${now}\:{checking}\:\:{between}\:\mathrm{1}\:{and}\:\mathrm{2} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{32}+\mathrm{48}+\mathrm{16}−\mathrm{64}−\mathrm{32}=\mathrm{0} \\ $$$${so}\:{x}=\mathrm{2}\:{is}\:{a}\:{root}.. \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}+\mathrm{6}+\mathrm{8}−\mathrm{4}−\mathrm{32}=−\mathrm{21} \\ $$$${now}\:{gradually}\:{we}\:{increase}\:{the}\:{value}\:{of}\:{x} \\ $$$${such}\:{as}\:\mathrm{1}.\mathrm{1},\mathrm{1}.\mathrm{2},...{and}\:{observe}\:{change}\:{of}\:{sign} \\ $$$${f}\left({x}\right)={y}=\left({x}^{\mathrm{5}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}\right)−\left(\mathrm{4}{x}^{\mathrm{4}} +\mathrm{32}\right) \\ $$$$\frac{{dy}}{{dx}}=\mathrm{5}{x}^{\mathrm{4}} +\mathrm{18}{x}^{\mathrm{2}} +\mathrm{8}−\mathrm{16}{x}^{\mathrm{3}} \\ $$$$\frac{\bigtriangleup{y}}{\bigtriangleup{x}}\approx\frac{{dy}}{{dx}} \\ $$$$\bigtriangleup{y}=\frac{{dy}}{{dx}}×\bigtriangleup{x}=\left(\mathrm{5}+\mathrm{18}+\mathrm{8}−\mathrm{16}\right)×\mathrm{0}.\mathrm{1}=\mathrm{1}.\mathrm{5} \\ $$$${so}\:{f}\left(\mathrm{1}.\mathrm{1}\right)={f}\left(\mathrm{1}\right)+\bigtriangleup{y}=−\mathrm{21}+\mathrm{1}.\mathrm{5}=−\mathrm{19}.\mathrm{5} \\ $$$${f}\left(\mathrm{2}.\mathrm{1}\right)={f}\left(\mathrm{2}\right)+\bigtriangleup{y}=\mathrm{0}+\mathrm{1}.\mathrm{5}=\mathrm{1}.\mathrm{5}>\mathrm{0} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{3}^{\mathrm{5}} +\mathrm{6}×\mathrm{3}^{\mathrm{3}} +\mathrm{8}×\mathrm{3}−\mathrm{4}×\mathrm{3}^{\mathrm{4}} −\mathrm{32} \\ $$$$=\mathrm{3}^{\mathrm{4}} \left(\mathrm{3}−\mathrm{4}\right)+\mathrm{6}×\mathrm{27}−\mathrm{8} \\ $$$$=−\mathrm{81}+\mathrm{162}−\mathrm{8}>\mathrm{0} \\ $$$${so}\:{no}\:{root}\:{between}\:\mathrm{2}.\mathrm{1}\:{and}\:\mathrm{3} \\ $$$$... \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{4}^{\mathrm{5}} +\mathrm{6}×\mathrm{4}^{\mathrm{3}} +\mathrm{32}−\mathrm{4}×\mathrm{4}^{\mathrm{4}} −\mathrm{32}>\mathrm{0} \\ $$$${so}\:{no}\:{root}\:{between}\:\mathrm{3}\:{and}\:\mathrm{4} \\ $$$${now}... \\ $$$${f}\left({x}\right)={x}^{\mathrm{5}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}−\left(\mathrm{4}{x}^{\mathrm{4}} +\mathrm{32}\right) \\ $$$${f}\left({x}\right)={g}\left({x}\right)−{h}\left({x}\right) \\ $$$$\frac{{df}}{{dx}}=\frac{{dg}}{{dx}}−\frac{{dh}}{{dx}}=\left(\mathrm{5}{x}^{\mathrm{4}} +\mathrm{18}{x}^{\mathrm{2}} +\mathrm{8}\right)−\left(\mathrm{16}{x}^{\mathrm{3}} \right) \\ $$$$={x}^{\mathrm{3}} \left(\mathrm{5}{x}−\mathrm{16}\right)+\mathrm{18}{x}^{\mathrm{2}} +\mathrm{8}.. \\ $$$${now}\:{look}\:{when}\:{x}>\mathrm{3}.\mathrm{2}\:\:\:\:\frac{{df}}{{dx}}>\mathrm{0} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{there}}\:\boldsymbol{{will}}\:\boldsymbol{{be}}\:\boldsymbol{{no}}\:\boldsymbol{{sign}}\:\boldsymbol{{change}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}>\mathrm{3}.\mathrm{2} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{this}}\:\boldsymbol{{eqn}}\:\boldsymbol{{has}}\:\boldsymbol{{only}}\:\boldsymbol{{one}}\:\boldsymbol{{root}}\:\boldsymbol{{that}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{x}}=\mathrm{2}...\boldsymbol{{pls}}\:\boldsymbol{{pay}}\:\boldsymbol{{your}}\:\boldsymbol{{kind}}\:\boldsymbol{{attention}}... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com