Question

find the roots of the quadratic equations 6x²-x-2=0(//////////only through complete square method/////////)​

Answer 1

       $\underline{\bold{Solution:}}$

       $\text{The given equation is }6x^{2} - x - 2 = 0$

       $\text{Dividing by 6}$

$\implies x^{2} - \dfrac{x}6 - \dfrac{1}{3} = 0$

       $\text{Applying the completing square method}$

$\implies x^{2} - \dfrac{x}{6} - \dfrac13 + (\dfrac{1}{12})^{2} - (\dfrac{1}{12})^{2} = 0$

       $\text{Rearranging the numbers}$

$\implies (x)^{2} - 2(\dfrac{1}{12})x + (\dfrac{1}{12})^{2} - (\dfrac{1}{12})^{2} - \dfrac13 = 0$

      $\text{We know that, }(a-b)^{2} = a^{2} + b^{2} - 2ab$

$\implies (x+\dfrac{1}{12})^{2} - \dfrac{1}{144} - \dfrac13 = 0$

       $\text{Simplifying...}$

$\implies (x+\dfrac{1}{12})^{2} + (\dfrac{-1-48}{144} )= 0$

       $\text{Simplifying...}$

$\implies (x+\dfrac{1}{12})^{2} - \dfrac{49}{144} = 0$

       $\text{We can write }\dfrac{49}{144}\text{ as }(\dfrac{7}{12})^{2}$

$\implies (x+\dfrac{1}{12})^{2} - (\dfrac{7}{12})^{2} = 0$

       $\text{We know that }a^{2} - b^{2} = (a+b)(a-b)$

$\implies (x+\dfrac{1}{12} + \dfrac{7}{12})(x+\dfrac{1}{12} - \dfrac{7}{12})= 0$

       $\text{Simplifying...}$

$\implies (x+\dfrac{8}{12})(x-\dfrac{6}{12})= 0$

       $\text{Simplifying...}$

$\implies (x+\dfrac{2}{3})(x-\dfrac{1}{2})= 0$

$\implies \boxed{x = -\dfrac{2}{3}, \dfrac12}$