Question

Find the roots of the following quadratic equation, if they exist using the quadratic formula of Sridharacharya: \[x+\dfrac{1}{x}=3,x\ne 0\].

Answer 1

- Hint: Solve the given expression until you get a simple expression of x. Now for x is an integer’s which comes in its range. Similarly, when x is a real number, find its range.

Complete step-by-step solution -

Shridhar Acharya formula is the quadratic formula, which is used for finding the roots of a quadratic equation, \[a{{x}^{2}}+bx+c=0\].
Where \[a\ne 0\] and a, b, c are real numbers and they are real coefficients of the equation.
Now, \[a{{x}^{2}}+bx+c=0\], has 2 roots which are,
\[x=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\] and \[x=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]
The above Shridhar Acharya’s rule can be proved by solving the general form of quadratic equation i.e. solving, \[a{{x}^{2}}+bx+c=0\].
Now we have been given the equation, \[x+\dfrac{1}{x}=3\].
Now let us make this equation, to a quadratic equation like, \[a{{x}^{2}}+bx+c=0\].
\[\begin{align}
  & x+\dfrac{1}{x}=3 \\
 & \dfrac{{{x}^{2}}+1}{x}=3\Rightarrow {{x}^{2}}+1=3x \\
 & \therefore {{x}^{2}}-3x+1=0 \\
\end{align}\]
Now the above equation is similar to \[a{{x}^{2}}+bx+c=0\].
Thus comparing them both we get,
a = 1, b = -3, c = 1.
Now let us put the above values on the 2 roots. We know that,
\[D={{b}^{2}}-4ac\]
Thus putting the values in the above expression, we get
\[D={{\left( -3 \right)}^{2}}-4\times 1\times 1=9-4=5\]
The root of the equation is given by, \[x=\dfrac{-b\pm \sqrt{D}}{2a}\].
Thus putting values,
\[x=\dfrac{-\left( -3 \right)\pm \sqrt{5}}{2\times 1}=\dfrac{3\pm \sqrt{5}}{2}\]
Thus we got the two roots as, \[x=\left( \dfrac{3+\sqrt{5}}{2} \right)\]and \[\left( \dfrac{3-\sqrt{5}}{2} \right)\].
Hence the roots of the quadratic equation \[\left( \dfrac{3+\sqrt{5}}{2} \right)\]and \[\left( \dfrac{3-\sqrt{5}}{2} \right)\].
Thus we got the required answer.

Note: The discriminant is the expression, \[{{b}^{2}}-4ac\], which is defined for any quadratic equation, \[a{{x}^{2}}+bx+c=0\]. Based upon the sign of expression, you can determine how many real number solutions that quadratic equation has. If D is positive then 2 unique solutions are there. If D = 0, one solution and D is negative then 2 imaginary roots.
Here D is positive, so 2 unique solutions \[\left( \dfrac{3+\sqrt{5}}{2} \right)\]and \[\left( \dfrac{3-\sqrt{5}}{2} \right)\].