Question
Solve for x.
$\text{x}+\frac{1}{\text{x}}=3,\text{x}\neq0$
$\text{x}+\frac{1}{\text{x}}=3,\text{x}\neq0$
✓
Answer
$\text{x}+\frac{1}{\text{x}}=3$
$\text{x}^2+1=3\text{x}$
$\text{x}^3-3\text{x}+1=0$
Here, $\text{a}=1,\text{b}=-3,\text{c}=1$
$\therefore\text{D}=\text{b}^2-4\text{ac}$
$=(-3)^2-4\times1\times1$
$=9-4=5$
$\because\text{D}>0$
$\therefore$ Roots are real
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-3)\pm\sqrt{5}}{2\times1}=\frac{(3)\pm\sqrt{5}}{2}$
$\therefore\text{x}=\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}$
$\text{x}^2+1=3\text{x}$
$\text{x}^3-3\text{x}+1=0$
Here, $\text{a}=1,\text{b}=-3,\text{c}=1$
$\therefore\text{D}=\text{b}^2-4\text{ac}$
$=(-3)^2-4\times1\times1$
$=9-4=5$
$\because\text{D}>0$
$\therefore$ Roots are real
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-3)\pm\sqrt{5}}{2\times1}=\frac{(3)\pm\sqrt{5}}{2}$
$\therefore\text{x}=\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}$
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