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 141794 by mathdanisur last updated on 23/May/21

Commented by JDamian last updated on 23/May/21

repeated question 141730

$${repeated}\:{question}\:\mathrm{141730} \\ $$

Commented by mathdanisur last updated on 23/May/21

solution, no answer

$${solution},\:{no}\:{answer} \\ $$

Answered by MJS_new last updated on 23/May/21

my path: let y=px∧z=qx (1) (p^2 +p+1)x^2 =9 (2) (p^2 +pq+q^2 )x^2 =16 (3) (q^2 +q+1)x^2 =25 ⇒ ((p^2 +p+1)/9)=((p^2 +pq+q^2 )/(16))=((q^2 +q+1)/(25)) (a)=(b)=(c) from solving (a)=(b) ∧ (b)=(c) we get q=((−p(2p+1))/(p−1)) this leads to (a)=(c) (p^2 +p+1)(p^2 +((32)/(11))p−((16)/(11)))=0 ⇒ p=−((16+12(√3))/(11))∨p=((−16+12(√3))/(11)) now insert backwards

$$\mathrm{my}\:\mathrm{path}: \\ $$$$\mathrm{let}\:{y}={px}\wedge{z}={qx} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right){x}^{\mathrm{2}} =\mathrm{9} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\left({p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} \right){x}^{\mathrm{2}} =\mathrm{16} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right){x}^{\mathrm{2}} =\mathrm{25} \\ $$$$\Rightarrow \\ $$$$\frac{{p}^{\mathrm{2}} +{p}+\mathrm{1}}{\mathrm{9}}=\frac{{p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} }{\mathrm{16}}=\frac{{q}^{\mathrm{2}} +{q}+\mathrm{1}}{\mathrm{25}} \\ $$$$\left({a}\right)=\left({b}\right)=\left({c}\right) \\ $$$$\mathrm{from}\:\mathrm{solving}\:\left({a}\right)=\left({b}\right)\:\wedge\:\left({b}\right)=\left({c}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$${q}=\frac{−{p}\left(\mathrm{2}{p}+\mathrm{1}\right)}{{p}−\mathrm{1}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\left({a}\right)=\left({c}\right) \\ $$$$\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)\left({p}^{\mathrm{2}} +\frac{\mathrm{32}}{\mathrm{11}}{p}−\frac{\mathrm{16}}{\mathrm{11}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{p}=−\frac{\mathrm{16}+\mathrm{12}\sqrt{\mathrm{3}}}{\mathrm{11}}\vee{p}=\frac{−\mathrm{16}+\mathrm{12}\sqrt{\mathrm{3}}}{\mathrm{11}} \\ $$$$\mathrm{now}\:\mathrm{insert}\:\mathrm{backwards} \\ $$

Answered by mr W last updated on 23/May/21

x^2 +y^2 +xy=9 ...(i) y^2 +z^2 +yz=16 ...(ii) z^2 +x^2 +zx=25 ...(iii) (ii)−(i): z^2 −x^2 +y(z−x)=7 (z−x)(x+y+z)=7 let s=x+y+z ⇒z=(7/s)+x (iii)−(ii): (x−y)(x+y+z)=9 ⇒y=−(9/s)+x x+y+z=x−(9/s)+x+(7/s)+x=3x−(2/s)=s ⇒x=(s/3)+(2/(3s)) ⇒y=−(9/s)+(s/3)+(2/(3s))=(s/3)−((25)/(3s)) insert into (i): ((s/3)+(2/(3s)))^2 +((s/3)−((25)/(3s)))^2 +((s/3)+(2/(3s)))((s/3)−((25)/(3s)))=9 s^2 −50+((193)/s^2 )=0 s^4 −50s^2 +193=0 ⇒s^2 =25±12(√3) (i)+(ii)+(iii): 2(x^2 +y^2 +z^2 )+(zy+yz+zx)=50 s^2 =(x+y+z)^2 =x^2 +y^2 +z^2 +2(xy+yz+zx) 2s^2 =2(x^2 +y^2 +z^2 )+xy+yz+zx+3(xy+yz+zx) 2(25±12(√3))=50+3(xy+yz+zx) ⇒xy+yz+zx=±8(√3)

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{9}\:\:\:\:\:...\left({i}\right) \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{yz}=\mathrm{16}\:\:\:\:...\left({ii}\right) \\ $$$${z}^{\mathrm{2}} +{x}^{\mathrm{2}} +{zx}=\mathrm{25}\:\:\:\:...\left({iii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$${z}^{\mathrm{2}} −{x}^{\mathrm{2}} +{y}\left({z}−{x}\right)=\mathrm{7} \\ $$$$\left({z}−{x}\right)\left({x}+{y}+{z}\right)=\mathrm{7} \\ $$$${let}\:{s}={x}+{y}+{z} \\ $$$$\Rightarrow{z}=\frac{\mathrm{7}}{{s}}+{x} \\ $$$$\left({iii}\right)−\left({ii}\right): \\ $$$$\left({x}−{y}\right)\left({x}+{y}+{z}\right)=\mathrm{9} \\ $$$$\Rightarrow{y}=−\frac{\mathrm{9}}{{s}}+{x} \\ $$$${x}+{y}+{z}={x}−\frac{\mathrm{9}}{{s}}+{x}+\frac{\mathrm{7}}{{s}}+{x}=\mathrm{3}{x}−\frac{\mathrm{2}}{{s}}={s} \\ $$$$\Rightarrow{x}=\frac{{s}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}{s}} \\ $$$$\Rightarrow{y}=−\frac{\mathrm{9}}{{s}}+\frac{{s}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}{s}}=\frac{{s}}{\mathrm{3}}−\frac{\mathrm{25}}{\mathrm{3}{s}} \\ $$$${insert}\:{into}\:\left({i}\right): \\ $$$$\left(\frac{{s}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}{s}}\right)^{\mathrm{2}} +\left(\frac{{s}}{\mathrm{3}}−\frac{\mathrm{25}}{\mathrm{3}{s}}\right)^{\mathrm{2}} +\left(\frac{{s}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}{s}}\right)\left(\frac{{s}}{\mathrm{3}}−\frac{\mathrm{25}}{\mathrm{3}{s}}\right)=\mathrm{9} \\ $$$${s}^{\mathrm{2}} −\mathrm{50}+\frac{\mathrm{193}}{{s}^{\mathrm{2}} }=\mathrm{0} \\ $$$${s}^{\mathrm{4}} −\mathrm{50}{s}^{\mathrm{2}} +\mathrm{193}=\mathrm{0} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\mathrm{25}\pm\mathrm{12}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right): \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+\left({zy}+{yz}+{zx}\right)=\mathrm{50} \\ $$$$ \\ $$$${s}^{\mathrm{2}} =\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$\mathrm{2}{s}^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+{xy}+{yz}+{zx}+\mathrm{3}\left({xy}+{yz}+{zx}\right) \\ $$$$\mathrm{2}\left(\mathrm{25}\pm\mathrm{12}\sqrt{\mathrm{3}}\right)=\mathrm{50}+\mathrm{3}\left({xy}+{yz}+{zx}\right) \\ $$$$\Rightarrow{xy}+{yz}+{zx}=\pm\mathrm{8}\sqrt{\mathrm{3}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com