Question

Find the root of the following equation
 \[\]$x - \dfrac{1}{x} = 3{\text{, }}x \ne 0$

Answer 1

Hint: Start by simplifying the equation by taking L.C.M. and then find out the value of Discriminant(${\text{ }}D = {b^2} - 4ac$) , by comparing the values of a, b ,c from the standard equation and look for the nature of roots and solve accordingly for the roots.

Complete step-by-step answer:
Given equation
$x - \dfrac{1}{x} = 3$
Let us start by simplifying the terms given.
Taking L.C.M as , we have
$\dfrac{{{x^2} - 1}}{x} = 3$
Cross-multiplying x to the side ,we get
${x^2} - 1 = 3x$
Shifting 3x to the other side
${x^2} - 1 - 3x = 0$
Let us compute the discriminant of this equation
We know, for an equation $a{x^2} + bx + c = 0$
$D = {b^2} - 4ac$
On comparison , we get
$
  a = 1,b = - 3,c = - 1 \\
  {\text{and }}D = {( - 3)^2} - 4(1)( - 1) \\
  D = 9 + 4 \\
  D = 13 \\
 $
D > 0 , So the roots are distinct and real.
The roots of the equation will be ${\text{ = }}\dfrac{{ - b \pm \sqrt D }}{{2a}}$
Substituting the values , we get
Roots ${\text{ = }}\dfrac{{ - ( - 3) \pm \sqrt {13} }}{2}{\text{ }}$
Therefore , the roots are $\dfrac{{3 + \sqrt {13} }}{2},\dfrac{{3 - \sqrt {13} }}{2}$

Note: Similar problems can be asked where the value of Discriminant will not be >0. In total there are three cases of Discriminant
D > 0 , Roots are real and distinct .
D = 0 , Roots are real and equal.
D < 0 , Roots are imaginary.
While substituting the values attention must be given to the coefficients of the terms (i.e. negative or positive sign).