Question Number 208681 by Noorzai last updated on 20/Jun/24
![](https://www.tinkutara.com/question/34174.png)
Answered by Rasheed.Sindhi last updated on 21/Jun/24
![(√(7+(√(48)))) =(√(7+4(√3))) =a+b(√3) (>0)(let) Where a,b∈Z 7+4(√3) =a^2 +3b^2 +2ab(√3) a^2 +3b^2 =7 & 2ab=4⇒b=(2/a) a^2 +3((2/a))^2 =7 a^2 +((12)/a^2 )=7 a^4 −7a^2 +12=0 (a^2 −3)(a^2 −4)=0 a^2 =4⇒a=±2 b=(2/(±2))=±1 a+b(√3) =2+(√3) , −2−(√3) (<0) ∴ a+b(√3) =2+(√3)](https://www.tinkutara.com/question/Q208684.png)
$$\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\:=\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}}\:={a}+{b}\sqrt{\mathrm{3}}\:\left(>\mathrm{0}\right)\left({let}\right) \\ $$$$\:{Where}\:{a},{b}\in\mathbb{Z} \\ $$$$\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\:={a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{\mathrm{3}}\: \\ $$$${a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} =\mathrm{7}\:\&\:\mathrm{2}{ab}=\mathrm{4}\Rightarrow{b}=\frac{\mathrm{2}}{{a}} \\ $$$${a}^{\mathrm{2}} +\mathrm{3}\left(\frac{\mathrm{2}}{{a}}\right)^{\mathrm{2}} =\mathrm{7} \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{12}}{{a}^{\mathrm{2}} }=\mathrm{7} \\ $$$${a}^{\mathrm{4}} −\mathrm{7}{a}^{\mathrm{2}} +\mathrm{12}=\mathrm{0} \\ $$$$\left({a}^{\mathrm{2}} −\mathrm{3}\right)\left({a}^{\mathrm{2}} −\mathrm{4}\right)=\mathrm{0} \\ $$$${a}^{\mathrm{2}} =\mathrm{4}\Rightarrow{a}=\pm\mathrm{2} \\ $$$${b}=\frac{\mathrm{2}}{\pm\mathrm{2}}=\pm\mathrm{1} \\ $$$${a}+{b}\sqrt{\mathrm{3}}\:=\mathrm{2}+\sqrt{\mathrm{3}}\:\:,\:−\mathrm{2}−\sqrt{\mathrm{3}}\:\left(<\mathrm{0}\right) \\ $$$$\therefore\:\:{a}+{b}\sqrt{\mathrm{3}}\:=\mathrm{2}+\sqrt{\mathrm{3}}\: \\ $$
Answered by a.lgnaoui last updated on 20/Jun/24
![=(√(7+4(√3))) =(√((4+3)+4(√3))) =(√((2+(√3) )^2 )) =2+(√3)](https://www.tinkutara.com/question/Q208682.png)
$$=\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$=\sqrt{\left(\mathrm{4}+\mathrm{3}\right)+\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$=\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} } \\ $$$$=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$