Question

How do you solve \[x + \dfrac{1}{x} = \dfrac{{10}}{3}\]?

Answer 1

Hint:In the left hand side of the equation we take LCM and simplify we will get a polynomial of degree 2. A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. If factors are difficult to find then we use Sridhar’s formula to find the roots.That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step by step answer:
Given, \[x + \dfrac{1}{x} = \dfrac{{10}}{3}\]
Taking LCM we have,
\[\dfrac{{{x^2} + 1}}{x} = \dfrac{{10}}{3}\]
Cross multiplying we have,
\[3\left( {{x^2} + 1} \right) = 10x\]
\[3{x^2} + 3 = 10x\]
Rearranging we have,
\[3{x^2} - 10x + 3 = 0\]
Thus we have a quadratic polynomial.
Since the degree of the equation is 2, we have 2 factors.
On comparing the given equation with the standard quadratic equation\[a{x^2} + bx + c = 0\], we have\[a = 3\], \[b = - 10\] and \[c = 3\].
The standard form of the factorization of quadratic equation is \[a{x^2} + {b_1}x + {b_2}x + c = 0\], which satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\].
We can write the given equation as \[3{x^2} - 9x - x + 3\], where \[{b_1} = - 9\] and \[{b_2} = - 1\]. Also \[{b_1} \times {b_2} = - 9 \times - 1 = 9(ac)\] and \[{b_1} + {b_2} = - 9 - 1 = - 10(b)\].
Thus we have,
\[ \Rightarrow 3{x^2} - 9x - x + 3 = 0\]
Taking ‘3x’ common in the first two terms and taking -1 common in the remaining two terms we have,
\[3x\left( {x - 3} \right) - 1(x - 3) = 0\]
Again taking \[\left( {x - 3} \right)\] common we have,
\[\left( {x - 3} \right)\left( {3x - 1} \right) = 0\]
By zero product principle we have,
\[\left( {x - 3} \right) = 0\] or \[\left( {3x - 1} \right) = 0\]
\[x = 3\] or \[3x = 1\]
\[x = 3\] and \[x = \dfrac{1}{3}\].

Hence the solutions are \[x = 3\] and \[x = \dfrac{1}{3}\].

Note: We can check if the obtained zeros are correct or not. Substitute \[x = 3\]in the given problem we have,
\[3 + \dfrac{1}{3} = \dfrac{{10}}{3}\]
Taking LCM and simplifying we have,
\[\dfrac{{9 + 1}}{3} = \dfrac{{10}}{3}\]
\[ \Rightarrow \dfrac{{10}}{3} = \dfrac{{10}}{3}\].
Similarly if we put \[x = \dfrac{1}{3}\] we get the same answer.
That is
\[\dfrac{1}{3} + \dfrac{1}{{\left( {\dfrac{1}{3}} \right)}} = \dfrac{{10}}{3}\]
\[\Rightarrow\dfrac{1}{3} + 3 = \dfrac{{10}}{3}\]
Taking LCM and simplifying we have,
\[\dfrac{{1 + 9}}{3} = \dfrac{{10}}{3}\]
\[ \therefore \dfrac{{10}}{3} = \dfrac{{10}}{3}\]. Hence the obtained answer is correct.