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 190093 by mnjuly1970 last updated on 27/Mar/23

       prove  that :    c= ( (√5) +2)^( (1/3))  − ((√5) −2)^( (1/3))               is   a  rational  number.

$$ \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$ \\ $$$${c}=\:\left(\:\sqrt{\mathrm{5}}\:+\mathrm{2}\right)^{\:\frac{\mathrm{1}}{\mathrm{3}}} \:−\:\left(\sqrt{\mathrm{5}}\:−\mathrm{2}\right)^{\:\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{is}\:\:\:\mathrm{a}\:\:{rational}\:\:\mathrm{number}. \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\: \\ $$

Commented by MJS_new last updated on 27/Mar/23

I don′t think it is rational

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is}\:\mathrm{rational} \\ $$

Commented by som(math1967) last updated on 18/May/23

yes sir, i think it is not rational

$${yes}\:{sir},\:{i}\:{think}\:{it}\:{is}\:{not}\:{rational} \\ $$

Commented by mnjuly1970 last updated on 27/Mar/23

yes sir thanks alot   i correted it.(√(5 )) instead of (√(10))

$${yes}\:{sir}\:{thanks}\:{alot} \\ $$$$\:{i}\:{correted}\:{it}.\sqrt{\mathrm{5}\:}\:{instead}\:{of}\:\sqrt{\mathrm{10}} \\ $$$$ \\ $$

Answered by MJS_new last updated on 27/Mar/23

...=(2+(√5))^(1/3) −(2+(√5))^(−1/3)   (2+(√5))^(1/3) =((1+(√5))/2)  ⇒ c=1

$$...=\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} −\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{−\mathrm{1}/\mathrm{3}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{c}=\mathrm{1} \\ $$

Answered by som(math1967) last updated on 27/Mar/23

 c^3 =((√5)+2)−((√5)−2)−3(5−4)^(1/3) .c  [c=((√5)+2)^(1/3) −((√5)−2)^(1/3) ]  c^3 =4−3c  c^3 +3c−4=0  c^3 −1+3(c−1)=0  (c−1)(c^2 +c+1+3)=0  (c−1)(c^2 +c+4)=0  for real value of c  (c^2 +c+4)≠0  c=1 ratinal number

$$\:{c}^{\mathrm{3}} =\left(\sqrt{\mathrm{5}}+\mathrm{2}\right)−\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)−\mathrm{3}\left(\mathrm{5}−\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} .{c} \\ $$$$\left[{c}=\left(\sqrt{\mathrm{5}}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right] \\ $$$${c}^{\mathrm{3}} =\mathrm{4}−\mathrm{3}{c} \\ $$$${c}^{\mathrm{3}} +\mathrm{3}{c}−\mathrm{4}=\mathrm{0} \\ $$$${c}^{\mathrm{3}} −\mathrm{1}+\mathrm{3}\left({c}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({c}−\mathrm{1}\right)\left({c}^{\mathrm{2}} +{c}+\mathrm{1}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\left({c}−\mathrm{1}\right)\left({c}^{\mathrm{2}} +{c}+\mathrm{4}\right)=\mathrm{0} \\ $$$${for}\:{real}\:{value}\:{of}\:{c}\:\:\left({c}^{\mathrm{2}} +{c}+\mathrm{4}\right)\neq\mathrm{0} \\ $$$${c}=\mathrm{1}\:{ratinal}\:{number} \\ $$

Commented by mehdee42 last updated on 27/Mar/23

that was perfect.

$${that}\:{was}\:{perfect}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com