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Find the roots of the equation: $x+\dfrac{1}{x}=3,x\ne 0$.

Last updated date: 16th Sep 2024
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Hint: We first simplify the equation to get quadratic form. Then we equate the given polynomial with the general form of the quadratic equation. We try to find the points where the curve intersects the X-axis. We take the x coordinates of those points using the theorem $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step-by-step solution
We need to find the roots of the equation $x+\dfrac{1}{x}=3,x\ne 0$.
We first simplify the equation to make the quadratic equation.
\begin{align} & x+\dfrac{1}{x}=3 \\ & \Rightarrow {{x}^{2}}+1=3x \\ & \Rightarrow {{x}^{2}}-3x+1=0 \\ \end{align}
We have a quadratic equation ${{x}^{2}}-3x+1=0$. Let $y\left( x \right)={{x}^{2}}-3x+1$.
We are finding the roots or zeros of the polynomial. The solutions are the points of x at which the polynomial value is 0. In the graphical form, we are finding the intersection points of the curve with the X-axis.
Now we verify it with the algebraic version of the solution.
We use the theorem $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for the general equation of polynomial $a{{x}^{2}}+bx+c=0$.
So, at those root points the equational value is 0. So, we are solving the equation $y\left( x \right)={{x}^{2}}-3x+1$. Here $a=1,b=-3,c=1$.
Putting values of $a=1,b=-3,c=1$ in the equation $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
$x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left\{ -\left( -3 \right) \right\}}^{2}}-4\times 1\times 1}}{2\times 1}=\dfrac{3\pm \sqrt{{{3}^{2}}-4\times 1\times 1}}{2\times 1}=\dfrac{3\pm \sqrt{5}}{2}$.
So, the roots of the equation $x+\dfrac{1}{x}=3$ are $x=\dfrac{3\pm \sqrt{5}}{2}$.

Note: We need to understand that the polynomial value has to be 0. Zeroes of the polynomial are the roots of the polynomial. So, at those points, the functional value of the curve is 0. The slope of the curve at those points is similar value-wise. We can also verify this result by substituting the values of zeros in the given equation ${{x}^{2}}-3x+2$ and check if the results satisfy or not. The $\sqrt{{{b}^{2}}-4ac}$ part in the form of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the determinant.