Question

Factorize $\sqrt{3}$ $x^2$ + 5x + 2$\sqrt{3}$ = 0 by splitting the middle term and find the value of x.

• A. $x=-\frac{2\sqrt{3}}{3},x=-\sqrt{3}$
• B. $x=\frac{2\sqrt{3}}{3},x=\sqrt{3}$
• C. $x=-\frac{2\sqrt{3}}{3},x=\sqrt{3}$
• D. $x=\frac{2\sqrt{3}}{3},x=-\sqrt{3}$

Answer 1

The correct option is A

$x=-\frac{2\sqrt{3}}{3},x=-\sqrt{3}$

Multiply $\sqrt{3}$ with 2$\sqrt{3}$ and we get 6

Factors of 6 are

1$\times$6 = 6 sum 1+ 6 = 7

2$\times$3= 6 sum 2+ 3 = 5

So our required pair of factors is 2 and 3

$\Rightarrow \sqrt{3}{x}^2+5x+2\sqrt{3}=0$

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

$\Rightarrow \sqrt{3}x(x+\sqrt{3})+2(x+\sqrt{3})=0$

$\Rightarrow (\sqrt{3}x+2)(x+\sqrt{3})=0$

$\therefore$ Factors are $(\sqrt{3}x+2),(x+\sqrt{3})$

$\Rightarrow (\sqrt{3}x+2)=0and(x+\sqrt{3})=0$

$\Rightarrow x=-\frac{2}{\sqrt{3}}andx=-\sqrt{3}$

Rationalising the denominator

$x=-\frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$x=-\frac{2\sqrt{3}}{3}$