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Question Number 173254 by mr W last updated on 09/Jul/22

solve for x,y,z ∈R  x+y+z=(√3)  xy+yz+zx=1

$${solve}\:{for}\:{x},{y},{z}\:\in{R} \\ $$$${x}+{y}+{z}=\sqrt{\mathrm{3}} \\ $$$${xy}+{yz}+{zx}=\mathrm{1} \\ $$

Commented by AgniMath last updated on 09/Jul/22

almost similar to my recent question

$$\mathrm{almost}\:\mathrm{similar}\:\mathrm{to}\:\mathrm{my}\:\mathrm{recent}\:\mathrm{question} \\ $$

Commented by mr W last updated on 09/Jul/22

yes. have you a solution other than   those below?

$${yes}.\:{have}\:{you}\:{a}\:{solution}\:{other}\:{than}\: \\ $$$${those}\:{below}? \\ $$

Commented by AgniMath last updated on 09/Jul/22

I know the same solution that is answered   by som(math 1967)

$$\mathrm{I}\:\mathrm{know}\:\mathrm{the}\:\mathrm{same}\:\mathrm{solution}\:\mathrm{that}\:\mathrm{is}\:\mathrm{answered}\: \\ $$$$\mathrm{by}\:\mathrm{som}\left(\mathrm{math}\:\mathrm{1967}\right) \\ $$

Commented by mr W last updated on 09/Jul/22

ok. i see.

$${ok}.\:{i}\:{see}. \\ $$

Answered by som(math1967) last updated on 09/Jul/22

(x+y+z)^2 =3  x^2 +y^2 +z^2 +2(xy+yz+zx)=3(xy+yz+zx)  [∵ xy+yz+zx=1]  ∴x^2 +y^2 +z^2 −xy−yz−zx=0  2x^2 +2y^2 +2z^2 −2xy−2yz−2zx=0  (x−y)^2 +(y−z)^2 +(z−x)^2 =0  ∵ x,y,z ∈R  ∴(x−y)^2 =0   (y−z)^2 =0   (z−x)^2 =0  ∴ x=y=z    x+y+z=(√3)    3x=(√3)   ∴ x=(1/( (√3)))   ∴x=y=z=(1/( (√3)))

$$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\mathrm{3} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right)=\mathrm{3}\left({xy}+{yz}+{zx}\right) \\ $$$$\left[\because\:{xy}+{yz}+{zx}=\mathrm{1}\right] \\ $$$$\therefore{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xy}−{yz}−{zx}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} −\mathrm{2}{xy}−\mathrm{2}{yz}−\mathrm{2}{zx}=\mathrm{0} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} +\left({y}−{z}\right)^{\mathrm{2}} +\left({z}−{x}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\because\:{x},{y},{z}\:\in{R} \\ $$$$\therefore\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{0}\:\:\:\left({y}−{z}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\left({z}−{x}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\therefore\:{x}={y}={z} \\ $$$$\:\:{x}+{y}+{z}=\sqrt{\mathrm{3}} \\ $$$$\:\:\mathrm{3}{x}=\sqrt{\mathrm{3}} \\ $$$$\:\therefore\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:\therefore{x}={y}={z}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$

Commented by mr W last updated on 09/Jul/22

very nice sir! thanks!

$${very}\:{nice}\:{sir}!\:{thanks}! \\ $$

Commented by Rasheed.Sindhi last updated on 09/Jul/22

Great sir!

$$\mathcal{G}{reat}\:{sir}! \\ $$

Commented by som(math1967) last updated on 09/Jul/22

My pleasure, good day sir

$${My}\:{pleasure},\:{good}\:{day}\:{sir} \\ $$

Answered by mr W last updated on 09/Jul/22

A still other way:  let y=px, z=qx  (p+q+1)x=(√3)  (p+q+pq)x^2 =1  (p+q+1)^2 =3(p+q+pq)  p^2 +q^2 +1=p+q+pq  (p−q)^2 +(p−1)(q−1)=0  (p−1−(q−1))^2 +(p−1)(q−1)=0  let P=p−1, Q=q−1  (P−Q)^2 +PQ=0 ⇒PQ=−(P−Q)^2 ≤0  P^2 +Q^2 −PQ=0 ⇒PQ=P^2 +Q^2 ≥0  ⇒PQ=0 ⇒P=Q=0  ⇒p−1=0, q−1=0   ⇒p=q=1  ⇒x=y=z=((√3)/3)

$${A}\:{still}\:{other}\:{way}: \\ $$$${let}\:{y}={px},\:{z}={qx} \\ $$$$\left({p}+{q}+\mathrm{1}\right){x}=\sqrt{\mathrm{3}} \\ $$$$\left({p}+{q}+{pq}\right){x}^{\mathrm{2}} =\mathrm{1} \\ $$$$\left({p}+{q}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3}\left({p}+{q}+{pq}\right) \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} +\mathrm{1}={p}+{q}+{pq} \\ $$$$\left({p}−{q}\right)^{\mathrm{2}} +\left({p}−\mathrm{1}\right)\left({q}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({p}−\mathrm{1}−\left({q}−\mathrm{1}\right)\right)^{\mathrm{2}} +\left({p}−\mathrm{1}\right)\left({q}−\mathrm{1}\right)=\mathrm{0} \\ $$$${let}\:{P}={p}−\mathrm{1},\:{Q}={q}−\mathrm{1} \\ $$$$\left({P}−{Q}\right)^{\mathrm{2}} +{PQ}=\mathrm{0}\:\Rightarrow{PQ}=−\left({P}−{Q}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$${P}^{\mathrm{2}} +{Q}^{\mathrm{2}} −{PQ}=\mathrm{0}\:\Rightarrow{PQ}={P}^{\mathrm{2}} +{Q}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow{PQ}=\mathrm{0}\:\Rightarrow{P}={Q}=\mathrm{0} \\ $$$$\Rightarrow{p}−\mathrm{1}=\mathrm{0},\:{q}−\mathrm{1}=\mathrm{0}\: \\ $$$$\Rightarrow{p}={q}=\mathrm{1} \\ $$$$\Rightarrow{x}={y}={z}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$

Commented by Rasheed.Sindhi last updated on 09/Jul/22

Great sir!

$$\mathbb{G}\boldsymbol{\mathrm{reat}}\:\boldsymbol{\mathrm{sir}}! \\ $$

Commented by Tawa11 last updated on 11/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 09/Jul/22

An other way:  x+y+z=(√3)  ⇒x+y=(√3)−z  xy+(x+y)z=1  xy+((√3)−z)z=1  ⇒xy=1+z^2 −(√3)z  x,y are roots of  t^2 +(z−(√3))t+(1+z^2 −(√3)z)=0  such that t, i.e.x,y exist,  Δ=(z−(√3))^2 −4(1+z^2 −(√3)z)≥0  3z^2 −2(√3)z+1≤0  ((√3)z−1)^2 ≤0  ⇒(√3)z−1=0  ⇒z=(1/( (√3)))=((√3)/3)  similarly  ⇒x=((√3)/3), y=((√3)/3)

$${An}\:{other}\:{way}: \\ $$$${x}+{y}+{z}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{x}+{y}=\sqrt{\mathrm{3}}−{z} \\ $$$${xy}+\left({x}+{y}\right){z}=\mathrm{1} \\ $$$${xy}+\left(\sqrt{\mathrm{3}}−{z}\right){z}=\mathrm{1} \\ $$$$\Rightarrow{xy}=\mathrm{1}+{z}^{\mathrm{2}} −\sqrt{\mathrm{3}}{z} \\ $$$${x},{y}\:{are}\:{roots}\:{of} \\ $$$${t}^{\mathrm{2}} +\left({z}−\sqrt{\mathrm{3}}\right){t}+\left(\mathrm{1}+{z}^{\mathrm{2}} −\sqrt{\mathrm{3}}{z}\right)=\mathrm{0} \\ $$$${such}\:{that}\:{t},\:{i}.{e}.{x},{y}\:{exist}, \\ $$$$\Delta=\left({z}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}+{z}^{\mathrm{2}} −\sqrt{\mathrm{3}}{z}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{z}+\mathrm{1}\leqslant\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{3}}{z}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$\Rightarrow\sqrt{\mathrm{3}}{z}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{z}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${similarly} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}},\:{y}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$

Commented by Rasheed.Sindhi last updated on 09/Jul/22

Great sir!

$$\mathbb{G}\boldsymbol{\mathrm{reat}}\:\boldsymbol{\mathrm{sir}}! \\ $$

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