please-solve-this-30-12-6-

Question Number 29909 by *D¬ B£$T* last updated on 13/Feb/18

$${please}\:{solve}\:{this}:\:\:\sqrt{\mathrm{30}+\mathrm{12}\sqrt{\mathrm{6}}} \\ $$

Commented by rahul 19 last updated on 13/Feb/18

(√(30+12(√(6 ))))= (√(30+2(√(216 ))=))((√a) +(√b)). so a+b=30 and ab=216 ⇒a=((216)/b). on solving we get a=18, b=12. hence it is equal to (√(18)) + (√(12)) . or we can say 3(√2) + 2(√3) .

$$\sqrt{\mathrm{30}+\mathrm{12}\sqrt{\mathrm{6}\:}}=\:\sqrt{\mathrm{30}+\mathrm{2}\sqrt{\mathrm{216}\:\:}=}\left(\sqrt{\mathrm{a}}\:+\sqrt{\mathrm{b}}\right). \\ $$$$\mathrm{so}\:\mathrm{a}+\mathrm{b}=\mathrm{30} \\ $$$$\mathrm{and}\:\mathrm{ab}=\mathrm{216} \\ $$$$\Rightarrow\mathrm{a}=\frac{\mathrm{216}}{\mathrm{b}}. \\ $$$$\mathrm{on}\:\mathrm{solving}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}=\mathrm{18},\:\mathrm{b}=\mathrm{12}. \\ $$$$\mathrm{hence}\:\mathrm{it}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\sqrt{\mathrm{18}}\:+\:\sqrt{\mathrm{12}}\:. \\ $$$$\mathrm{or}\:\mathrm{we}\:\mathrm{can}\:\mathrm{say}\:\mathrm{3}\sqrt{\mathrm{2}}\:+\:\mathrm{2}\sqrt{\mathrm{3}}\:. \\ $$

Answered by mrW2 last updated on 13/Feb/18

let 30+12(√6)=(p+q(√6))^2 =p^2 +2pq(√6)+6q^2 2pq=12⇒pq=6 p^2 +6q^2 =30 ⇒p^2 +6((6/p))^2 =30 ⇒p^4 −30p^2 +216=0 ⇒p^2 =((30±(√(30^2 −4×216)))/2)=((30±6)/2)= { ((18)),((12)) :} ⇒p= { ((3(√2))),((2(√3))) :} ⇒q= { (((6/(3(√2)))=(√2))),(((6/(2(√3)))=(√3))) :} ⇒(√(30+12(√6)))=p+q(√6)=3(√2)+(√2)×(√6)=3(√2)+2(√3) or ⇒(√(30+12(√6)))=2(√3)+(√3)×(√6)=2(√3)+3(√2)=the same as abov

$${let}\:\mathrm{30}+\mathrm{12}\sqrt{\mathrm{6}}=\left({p}+{q}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +\mathrm{2}{pq}\sqrt{\mathrm{6}}+\mathrm{6}{q}^{\mathrm{2}} \\ $$$$\mathrm{2}{pq}=\mathrm{12}\Rightarrow{pq}=\mathrm{6} \\ $$$${p}^{\mathrm{2}} +\mathrm{6}{q}^{\mathrm{2}} =\mathrm{30} \\ $$$$\Rightarrow{p}^{\mathrm{2}} +\mathrm{6}\left(\frac{\mathrm{6}}{{p}}\right)^{\mathrm{2}} =\mathrm{30} \\ $$$$\Rightarrow{p}^{\mathrm{4}} −\mathrm{30}{p}^{\mathrm{2}} +\mathrm{216}=\mathrm{0} \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\frac{\mathrm{30}\pm\sqrt{\mathrm{30}^{\mathrm{2}} −\mathrm{4}×\mathrm{216}}}{\mathrm{2}}=\frac{\mathrm{30}\pm\mathrm{6}}{\mathrm{2}}=\begin{cases}{\mathrm{18}}\\{\mathrm{12}}\end{cases} \\ $$$$\Rightarrow{p}=\begin{cases}{\mathrm{3}\sqrt{\mathrm{2}}}\\{\mathrm{2}\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\Rightarrow{q}=\begin{cases}{\frac{\mathrm{6}}{\mathrm{3}\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}}\\{\frac{\mathrm{6}}{\mathrm{2}\sqrt{\mathrm{3}}}=\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\Rightarrow\sqrt{\mathrm{30}+\mathrm{12}\sqrt{\mathrm{6}}}={p}+{q}\sqrt{\mathrm{6}}=\mathrm{3}\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}×\sqrt{\mathrm{6}}=\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${or} \\ $$$$\Rightarrow\sqrt{\mathrm{30}+\mathrm{12}\sqrt{\mathrm{6}}}=\mathrm{2}\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}×\sqrt{\mathrm{6}}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{2}}={the}\:{same}\:{as}\:{abov} \\ $$

Commented by NECx last updated on 13/Feb/18

$${so}\:{nice}\:{workings} \\ $$

Answered by ajfour last updated on 13/Feb/18

let a=(√(30+12(√6))) ⇒ a^2 =30+12(√6) .....(i) (1/a^2 )=(1/(30+12(√6))) = ((30−12(√6))/(36)) or ((36)/a^2 ) = 30−12(√6) ...(ii) (i)+(ii) gives: a^2 +((36)/a^2 ) = 60 ⇒ (a+(6/a))^2 =60+12 ⇒ a+(6/a) =6(√2) or a^2 −6(√2)a+6=0 (a−3(√2))^2 = 12 a = 2(√3)+3(√2) .

$${let}\:\:{a}=\sqrt{\mathrm{30}+\mathrm{12}\sqrt{\mathrm{6}}} \\ $$$$\Rightarrow\:\:{a}^{\mathrm{2}} =\mathrm{30}+\mathrm{12}\sqrt{\mathrm{6}}\:\:\:…..\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{30}+\mathrm{12}\sqrt{\mathrm{6}}}\:=\:\frac{\mathrm{30}−\mathrm{12}\sqrt{\mathrm{6}}}{\mathrm{36}} \\ $$$${or}\:\:\:\:\frac{\mathrm{36}}{{a}^{\mathrm{2}} }\:=\:\mathrm{30}−\mathrm{12}\sqrt{\mathrm{6}}\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)\:{gives}: \\ $$$$\:\:\:\:\:\:\:{a}^{\mathrm{2}} +\frac{\mathrm{36}}{{a}^{\mathrm{2}} }\:=\:\mathrm{60} \\ $$$$\Rightarrow\:\:\:\:\left({a}+\frac{\mathrm{6}}{{a}}\right)^{\mathrm{2}} =\mathrm{60}+\mathrm{12}\: \\ $$$$\Rightarrow\:\:\:\:\:\:{a}+\frac{\mathrm{6}}{{a}}\:=\mathrm{6}\sqrt{\mathrm{2}} \\ $$$${or}\:\:\:\:\:\:\:{a}^{\mathrm{2}} −\mathrm{6}\sqrt{\mathrm{2}}{a}+\mathrm{6}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({a}−\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{a}}\:=\:\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{2}}\:\:. \\ $$

Commented by 33 last updated on 13/Feb/18