Question

$3x {}^{2} - 4 \sqrt{3x} + 4 = 0$
​

Answer 1

$\huge{ \green{ \mathfrak{ \underline{Question}}}}$

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

■☆$Factorize:\; 3x {}^{2} - 4 \sqrt{3}x + 4 = 0$☆■

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

$\huge{ \green{ \mathfrak{ \underline{Answer}}}}$

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

$\boxed{\red{\bold{Solving \: by \: splitting\: terms:}}}$

$3x^2 - 4\sqrt{3}x +4 = 0$

$3x^2 - 2\sqrt{3}x -2\sqrt{3}x + 4 = 0$

$\blue{\sqrt{3}x(\sqrt{3}x-2) -2(\sqrt{3}x - 2)} = 0$

$\boxed{\red{\bold{By \: grouping \: method:}}}$

$\purple{(\sqrt{3}x -2)(\sqrt{3}x - 2)=0}$

$\boxed{\red{\bold{A \: trick\: used:}}}$

■☆Just divide the product to be seen by the root value and then you can simply factorize it. ☆■

■☆For example , in this case , the product was 12 ; and the root is of 3 , so dividing the product by 3 , you get 4 . This means that sum is 4 , as coefficient of √3 is 4 , and product is also 4 . Thus it is 2 . and thats how I factorized it as 2√3☆■

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂