Question-198754

Question Number 198754 by essaad last updated on 24/Oct/23

Answered by Frix last updated on 24/Oct/23

x^(1/2) +x^(2/3) =12 We can only approximate x≈20.4490371248

$${x}^{\frac{\mathrm{1}}{\mathrm{2}}} +{x}^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{12} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}\approx\mathrm{20}.\mathrm{4490371248} \\ $$

Commented by ajfour last updated on 24/Oct/23

Let x^(1/6) =t t^3 +t^4 =12 let (t^2 +ht+p)(t^2 +kt+q)=0 h+k=1 p+q+hk=0 (h+k)(p+q)−(h−k)(p−q)=0 ...(i) ⇒ pq=−12 hence h^2 k^2 =(1−4hk)(h^2 k^2 +48) say hk=z z^3 +48z−12=0 z=(2(√(1033))+6)^(1/3) −(2(√(1033))−6)^(1/3) z≈0.249675744 t=−(h/2)±(√((h^2 /4)−p)) p+q=−z pq=−12 ⇏ p, q =−(z/2)±(√((z^2 /4)+12)) now from ...(i) let h>k ; then h−k=(√((h+k)^2 −4hk)) ⇒ h−k=(√(1−4z)) h+k=1 ⇒ 2h=1+(√(1−4z)) >0 if h∈R 2k=1−(√(1−4z)) If x>0 then t>0 ⇒ p=−(√((z^2 /4)+12))−(z/2) ( <0 ) t=x^(1/6) =(√((h^2 /4)−p))−((h/2)) x^(1/6) = (√((((1+(√(1−4z)))^2 )/(16))+(√((z^2 /4)+12))+(z/2)))−(((1+(√(1−4z)))/4)) where z=(2(√(1033))+6)^(1/3) −(2(√(1033))−6)^(1/3) z ≈ 0.249675744 ((1+(√(1−4z)))/4)≈ 0.259003555 x^(1/6) ≈ 1.65368425 x≈ 20.45

$${Let}\:\:{x}^{\mathrm{1}/\mathrm{6}} ={t} \\ $$$${t}^{\mathrm{3}} +{t}^{\mathrm{4}} =\mathrm{12} \\ $$$${let} \\ $$$$\left({t}^{\mathrm{2}} +{ht}+{p}\right)\left({t}^{\mathrm{2}} +{kt}+{q}\right)=\mathrm{0} \\ $$$${h}+{k}=\mathrm{1} \\ $$$${p}+{q}+{hk}=\mathrm{0} \\ $$$$\left({h}+{k}\right)\left({p}+{q}\right)−\left({h}−{k}\right)\left({p}−{q}\right)=\mathrm{0}\:\:\:\:…\left({i}\right) \\ $$$$\Rightarrow\:\: \\ $$$${pq}=−\mathrm{12} \\ $$$${hence} \\ $$$${h}^{\mathrm{2}} {k}^{\mathrm{2}} =\left(\mathrm{1}−\mathrm{4}{hk}\right)\left({h}^{\mathrm{2}} {k}^{\mathrm{2}} +\mathrm{48}\right) \\ $$$${say}\:\:\:{hk}={z} \\ $$$${z}^{\mathrm{3}} +\mathrm{48}{z}−\mathrm{12}=\mathrm{0} \\ $$$${z}=\left(\mathrm{2}\sqrt{\mathrm{1033}}+\mathrm{6}\right)^{\mathrm{1}/\mathrm{3}} −\left(\mathrm{2}\sqrt{\mathrm{1033}}−\mathrm{6}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${z}\approx\mathrm{0}.\mathrm{249675744} \\ $$$${t}=−\frac{{h}}{\mathrm{2}}\pm\sqrt{\frac{{h}^{\mathrm{2}} }{\mathrm{4}}−{p}} \\ $$$${p}+{q}=−{z} \\ $$$${pq}=−\mathrm{12} \\ $$$$\nRightarrow\:\:{p},\:{q}\:=−\frac{{z}}{\mathrm{2}}\pm\sqrt{\frac{{z}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{12}} \\ $$$${now}\:{from}\:\:…\left({i}\right) \\ $$$${let}\:\:\:\:{h}>{k}\:\:;\:{then} \\ $$$${h}−{k}=\sqrt{\left({h}+{k}\right)^{\mathrm{2}} −\mathrm{4}{hk}} \\ $$$$\Rightarrow\:\:{h}−{k}=\sqrt{\mathrm{1}−\mathrm{4}{z}} \\ $$$$\:\:\:\:\:\:\:{h}+{k}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\mathrm{2}{h}=\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{4}{z}}\:\:\:\:\:>\mathrm{0}\:\:\:{if}\:{h}\in\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}{k}=\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{4}{z}} \\ $$$${If}\:\:\:{x}>\mathrm{0}\:\:{then}\:\:\:{t}>\mathrm{0} \\ $$$$\Rightarrow\:\:{p}=−\sqrt{\frac{{z}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{12}}−\frac{{z}}{\mathrm{2}}\:\:\:\:\:\left(\:\:<\mathrm{0}\:\right) \\ $$$${t}={x}^{\mathrm{1}/\mathrm{6}} =\sqrt{\frac{{h}^{\mathrm{2}} }{\mathrm{4}}−{p}}−\left(\frac{{h}}{\mathrm{2}}\right) \\ $$$$\:\:\:{x}^{\mathrm{1}/\mathrm{6}} = \\ $$$$\sqrt{\frac{\left(\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{4}{z}}\right)^{\mathrm{2}} }{\mathrm{16}}+\sqrt{\frac{{z}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{12}}+\frac{{z}}{\mathrm{2}}}−\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{4}{z}}}{\mathrm{4}}\right) \\ $$$$\:{where}\:{z}=\left(\mathrm{2}\sqrt{\mathrm{1033}}+\mathrm{6}\right)^{\mathrm{1}/\mathrm{3}} −\left(\mathrm{2}\sqrt{\mathrm{1033}}−\mathrm{6}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${z}\:\approx\:\mathrm{0}.\mathrm{249675744} \\ $$$$\frac{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{4}{z}}}{\mathrm{4}}\approx\:\mathrm{0}.\mathrm{259003555} \\ $$$${x}^{\mathrm{1}/\mathrm{6}} \approx\:\mathrm{1}.\mathrm{65368425} \\ $$$${x}\approx\:\mathrm{20}.\mathrm{45} \\ $$

Commented by Tawa11 last updated on 24/Oct/23

$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$

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