Question

Find the roots of the following quadratic equation by factorization $4{x^2} - 12x + 9 = 0$ and $4\sqrt {3{x^2}} + 5x - 2\sqrt 3 = 0$

Answer 1

Hint: In order to solve this according to the factorization method we have to break the middle term. In the first equation split the middle term i.e. $12x$ as $\left( {6 + 6} \right)x$ and in second equation i.e.$5x$ as $\left( {8 - 3} \right)x$.

Complete step-by-step answer:

Taking the first equation:
$4{x^2} - 12x + 9 = 0$
first of all split the middle term we get:
$
   \Rightarrow 4{x^2} - \left( {6 + 6} \right)x + 9 = 0 \\
   \Rightarrow 4{x^2} - 6x - 6x + 9 = 0 \\
   \Rightarrow 2x\left( {2x - 3} \right) - 3\left( {2x - 3} \right) = 0 \\
   \Rightarrow \left( {2x - 3} \right)\left( {2x - 3} \right) = 0 \\
$

$ \Rightarrow \left( {2x - 3} \right) = 0$ and $\left( {2x - 3} \right) = 0$
$\therefore x = \dfrac{3}{2}$ and $x = \dfrac{3}{2}$

Now taking the second equation given in question:
$4\sqrt {3{x^2}} + 5x - 2\sqrt 3 = 0$
now split the middle term we get:
$
   \Rightarrow 4\sqrt {3{x^2}} + \left( {8 - 3} \right)x - 2\sqrt 3 = 0 \\
   \Rightarrow 4\sqrt {3{x^2}} + 8x - 3x - 2\sqrt 3 = 0 \\
   \Rightarrow 4x\left( {\sqrt 3 x + 2} \right) - \sqrt 3 \left( {\sqrt 3 x + 2} \right) = 0 \\
   \Rightarrow \left( {\sqrt 3 x + 2} \right)\left( {4x - \sqrt 3 } \right) = 0 \\
$
$ \Rightarrow \sqrt 3 x + 2 = 0$ and $\left( {4x - \sqrt 3 } \right) = 0$
$\therefore x = \dfrac{{ - 2}}{{\sqrt 3 }}$ and $x = \dfrac{{\sqrt 3 }}{4}$

Note: In factorization we represent a given polynomial equation as a product of two or more terms, so in this question first we took the first equation and split the middle term after that we took $2x$ and $ - 3$ common and simplified the equation and got the value of $x$, similarly in equation second after splitting the middle term we took $4x$ and $ - \sqrt 3 $ common and simplified the equation and got the value of $x$.