Question

Under root3x^2+5x-2root3.. plzzzzz solve it my loving friendsss..... plzzzzzz

Answer 1

Given:

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

To find:

solve=?

Solution:

Formula:

$\bold{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$

compare the given value by:

$\sqrt{3}x^2+5x-2\sqrt{3}=0\\\\ax^2+bx+c=0\\\\\to a= \sqrt{3}\\\to b= 5 \\\to c= -2 \sqrt{3}$

apply the value in the above-given formula:

$\Rightarrow x=\frac{-5\pm\sqrt{5^2-4\times \sqrt{3} \times(-2\sqrt{3})}}{2\times \sqrt{3}}$

$\Rightarrow x=\frac{-5\pm\sqrt{25+8\times 3}}{2\times \sqrt{3}}\\\\\Rightarrow x=\frac{-5\pm\sqrt{25+24}}{ 2\sqrt{3}}\\\\\Rightarrow x=\frac{-5\pm\sqrt{49}}{ 2\sqrt{3}}\\\\\Rightarrow x=\frac{-5\pm7}{ 2\sqrt{3}}$

$\Rightarrow x= \frac{-5+7}{2\sqrt{3}} \ \ \ \ \ \ and \ \ \ \ x= \frac{-5-7}{2\sqrt{3}}\\\\\Rightarrow x= \frac{2}{2\sqrt{3}} \ \ \ \ \ \ and \ \ \ \ x= \frac{-12}{2\sqrt{3}}\\\\\Rightarrow x= \frac{1}{\sqrt{3}} \ \ \ \ \ \ and \ \ \ \ x= \frac{-6}{\sqrt{3}}\\\\\Rightarrow x= \frac{1}{\sqrt{3}} \ \ \ \ \ \ and \ \ \ \ x= \frac{-6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\\\\\Rightarrow x= \frac{1}{\sqrt{3}} \ \ \ \ \ \ and \ \ \ \ x= \frac{-6\times \sqrt{3}}{3}$

$\Rightarrow x= \frac{1}{\sqrt{3}} \ \ \ \ \ \ and \ \ \ \ x= -2\times \sqrt{3}$