Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Solve $30({x^2} + \dfrac{1}{{{x^2}}}) - 77(x - \dfrac{1}{x}) - 12 = 0$

Last updated date: 17th Jun 2024
Total views: 393.3k
Views today: 10.93k
Verified
393.3k+ views
Hint: For easy and convenient mean of solving the above equation we need to find a link between $({x^2} + \dfrac{1}{{{x^2}}})$ and $(x - \dfrac{1}{x})$ . Once we get a link we can replace the common terms by a single letter such as ‘m’ or ‘n’ except ‘x’. Then using the middle term factorization method we need to find the solution of the derived equation and from there again using the same middle term factorization method we will get to know the values of x.

Given:
$30({x^2} + \dfrac{1}{{{x^2}}}) - 77(x - \dfrac{1}{x}) - 12 = 0$
We have to find the possible values of x i.e. the roots of the given equation.
$30({x^2} + \dfrac{1}{{{x^2}}}) - 77(x - \dfrac{1}{x}) - 12 = 0$
Now, ${(x - \dfrac{1}{x})^2} = {x^2} + \dfrac{1}{{{x^2}}} - 2$
Hence, ${x^2} + \dfrac{1}{{{x^2}}} = {(x - \dfrac{1}{x})^2} + 2$
Now, putting the above value ${x^2} + \dfrac{1}{{{x^2}}} = {(x - \dfrac{1}{x})^2} + 2$ in the given equation, we will get as below:
$30({(x - \dfrac{1}{x})^2} + 2) - 77(x - \dfrac{1}{x}) - 12 = 0$
Now, let $(x - \dfrac{1}{x}) = m$ , putting the value $(x - \dfrac{1}{x}) = m$ in the above equation, we will get as below:
$\Rightarrow 30({m^2} + 2) - 77m - 12 = 0$
$\Rightarrow 30{m^2} + 60 - 77m - 12 = 0$
$\Rightarrow 30{m^2} - 77m + 48 = 0$
Now, we need to factorize the derived equation using the middle term factorization method
So, we will be getting as below
$\Rightarrow 30{m^2} - 45m - 32m + 48 = 0$
$\Rightarrow 15m(2m - 3) - 16(2m - 3) = 0$
$\Rightarrow (15m - 16)(2m - 3) = 0$
So, $15m - 16 = 0$
Therefore, $m = \dfrac{{16}}{{15}}$
Similarly, $m = \dfrac{3}{2}$
Now, substituting the value of m since $(x - \dfrac{1}{x}) = m$ we will be getting as
$\Rightarrow (x - \dfrac{1}{x}) = \dfrac{{16}}{{15}}$ and $(x - \dfrac{1}{x}) = \dfrac{3}{2}$
Now, lets solve $(x - \dfrac{1}{x}) = \dfrac{{16}}{{15}}$
$\Rightarrow (x - \dfrac{1}{x}) = \dfrac{{16}}{{15}}$
$\Rightarrow \dfrac{{{x^2} - 1}}{x} = \dfrac{{16}}{{15}}$
$\Rightarrow 15({x^2} - 1) = 16x$
$\Rightarrow 15{x^2} - 16x - 15 = 0$
$\Rightarrow 15{x^2} - 25x + 9x - 15 = 0$
$\Rightarrow 5x(3x - 5) + 3(3x - 5) = 0$
$(5x + 3)(3x - 5) = 0$
So, $(5x + 3) = 0$
Hence, $x = - \dfrac{3}{5}$
Similarly, $(3x - 5) = 0$
Hence, $x = \dfrac{5}{3}$
Now, lets solve $(x - \dfrac{1}{x}) = \dfrac{3}{2}$
$(x - \dfrac{1}{x}) = \dfrac{3}{2}$
$\dfrac{{{x^2} - 1}}{x} = \dfrac{3}{2}$
$\Rightarrow 2({x^2} - 1) = 3x$
$\Rightarrow 2{x^2} - 3x - 2 = 0$
$\Rightarrow 2{x^2} - 4x + 1x - 2 = 0$
$\Rightarrow 2x(x - 2) + 1(x - 2) = 0$
$\Rightarrow (2x + 1)(x - 2) = 0$
So, $(2x + 1) = 0$
Hence, $x = - \dfrac{1}{2}$
Similarly, $(x - 2) = 0$
Hence, $x = 2$
Therefore the possible values of x or we can say the roots of the given equation are $- \dfrac{3}{5}$ , $\dfrac{5}{3}$ , $- \dfrac{1}{2}$ , $2$ .
So, the correct answer is “ $- \dfrac{3}{5}$ , $\dfrac{5}{3}$ , $- \dfrac{1}{2}$ , $2$ .”.

Note: Though the above process seems simple but yet sometimes it can create confusion because we are doing substitution and also implementing the middle term factorization method number of times. So, be careful and don’t miss the values of x. Remember, in case of ${x^2}$ or $\dfrac{1}{{{x^2}}}$ there will be four roots and in case of x or $\dfrac{1}{x}$ there will be two roots.