Question

$find \: the \: of \: x - \frac{3}{x } = 2$
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Answer 1

Question :-

$\sf \: find \: the \: value \: of \: x - \frac{3}{x} = 2$

Answer :-

$\longmapsto \boxed{ \: \sf x = 3 \: or \: - 1 }$

Solution :-

We have ,

$\to \sf \: x - \frac{3}{x} = 2\\ \\ \bf \: taking \: lcm \: \\ \to \sf \: \frac{ {x }^{2} - 3}{x} = 2 \\ \\ \to \sf \: {x}^{2} - 3 = 2x \\ \\ \to \sf \: {x}^{2} - 2x - 3 = 0 \\ \\ \text{solve \: this \: quadratic \: by \: middle \: term } \\ \text{ splitting \: method} \\ \\ \to \sf \: {x}^{2} - 3x + x - 3 = 0 \\ \\ \to \sf \: x(x - 3) + 1(x - 3) = 0 \\ \\ \to \sf \: (x - 3)(x + 1) = 0 \\ \\ \star \sf \: case \: (1) \: if \: \\ \to \sf \: x - 3 = 0 \\ \\ \to \sf \boxed{ \sf \: x = 3} \\ \\ \star \sf \: case \: (2) \: if \\ \\ \to \sf \: x + 1 = 0 \\ \\ \to \boxed{ \sf \: x = 1} \\ \\ \sf hence \: \: x = 3 \: or \: - 1$

$\underline{ \boxed{\sf verification \: }} : - \\ \\ \bf \: let \\ \sf f(x) = {x}^{2} - 2x - 3 = 0 \\ \\ \sf \: when \: x = 3 \\ \\ \to \sf f(3) = {3}^{2} - 3 \times 2 - 3 = 0 \\ \\ \to \: 9 - 6 - 3 = 0 \\ \\ \to \: 0 = 0 \\ \\ \sf \: when \: x = - 1 \\ \\ \to \sf \: f( - 1) = {( - 1)}^{2} - 2 \times ( - 1) - 3 = 0 \\ \\ \to \: 1 + 2 - 3 = 0 \\ \\ \to \: 0 = 0 \\ \\ \large \sf \: hence \: verified \:$