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प्रश्न
Using properties of proportion solve:
\[ \Rightarrow \frac{x^2-x+1}{x^2+x+1}=\frac{112(1-x)}{104(1+x)} \]
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उत्तर
Given,
\[ \Rightarrow \frac{x^2-x+1}{x^2+x+1}=\frac{112(1-x)}{104(1+x)} \]
\[ \Rightarrow \frac{x^2-x+1}{x^2+x+1}=\frac{14(1-x)}{13(1+x)} \]
Applying componendo and dividendo we get,
\[ \Rightarrow \frac{(x^2-x+1)+(x^2+x+1)} {(x^2-x+1)-(x^2+x+1)} = \frac{14(1-x)+13(1+x)} {14(1-x)-13(1+x)} \]
\[ \Rightarrow \frac{x^2-x+1+x^2+x+1} {x^2-x+1-x^2-x-1} = \frac{14-14x+13+13x} {14-14x-13-13x} \]
\[ \Rightarrow \frac{2x^2+2}{-2x} = \frac{27-x}{1-27x} \]
\[ \Rightarrow \frac{2(x^2+1)}{-2x} = \frac{27-x}{1-27x} \]
\[ \Rightarrow \frac{x^2+1}{-x} = \frac{27-x}{1-27x} \]
\[ \Rightarrow (x^2+1)(1-27x)=-x(27-x) \]
\[ \Rightarrow x^2+1-27x^3-27x=-27x+x^2 \]
\[ \Rightarrow -27x^3+1=0 \]
\[ \Rightarrow 27x^3=1 \]
\[ \Rightarrow x^3=\frac{1}{27} \]
\[ \Rightarrow x=\sqrt[3]{\frac{1}{27}}=\frac{1}{3} \]
