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Question Number 173242 by AgniMath last updated on 08/Jul/22

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

$$2\mathit{a},\mathit{b},\mathit{c}\in \mathit{R}$$

Answered by AgniMath last updated on 09/Jul/22

$$\mathrm{Another}\mathrm{way}$$$$$$$$(a+b+c)=\sqrt{3}$$$$or{(a+b+c)}^2=3$$$$or{a}^2+b^2+c^2+2ab+2bc+2ca=3$$$$or{a}^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca$$$$[\because ab+bc+ca=1]$$$$or{a}^2+b^2+c^2-ab-bc-ca=0$$$$or2a^2+2b^2+2c^2-2ab-2bc-2ca=0$$$$or{(a-b)}^2+{(b-c)}^2+{(c-a)}^2=0$$$$$$$$\left|\begin{array}{c}{(a-b)}^2=0\\ ora-b=0\\ ora=b\end{array}\right|\left|\begin{array}{c}{(b-c)}^2=0\\ orb-c=0\\ b=c\end{array}\right|\left|\begin{array}{c}{(c-a)}^2=0\\ orc-a=0\\ orc=a\end{array}\right|$$$$\therefore a=b=c$$$$$$$$\frac{a+b}{c}=\frac{a+a}{a}=\frac{2\cancel{a}}{\cancel{a}}=2$$

Answered by a.lgnaoui last updated on 09/Jul/22

$$a+b+c=\sqrt{3}\Rightarrow \frac{a+b}{c}+1=\frac{\sqrt{3}}{c}$$$$ab+bc+ca=1\Rightarrow ab+c(a+b)=1\Rightarrow ab=1-c(a+b)$$$$\frac{ab}{c}+(a+b)=\frac{1}{c}ab=1-c(a+b)$$$$\frac{a+b}{c}+1=\frac{\sqrt{3}}{c}a+b=\sqrt{3}-c$$$$ab=1-(\sqrt{3}-c)c=1-\sqrt{3}c+c^2$$$$x^2-(\sqrt{3}-c)x+c^2-\sqrt{3}c+1$$$$3+c^2-2\sqrt{3}c-4(c^2-\sqrt{3}c+1)=3-4-3c^2+2\sqrt{3}c$$$$3c^2-2\sqrt{3}c+1\Rightarrow {(c-\frac{1}{3})}^2$$$$c=\frac{2\sqrt{3}-}{6}=\frac{\sqrt{3}}{3}\Rightarrow a+b=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$$$$\frac{a+b}{c}=\frac{2\sqrt{3}}{3}\times \frac{3}{\sqrt{3}}=2isthepossiblesolution$$

Commented by Tawa11 last updated on 11/Jul/22

$$\mathrm{Great}\mathrm{sirs}$$

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