Question

$\large\boxed{\fcolorbox{red}{orange}{Question}}$

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

√3x²-2√2x-2√3=0​

Answer 1

Answer :-

The roots of x in the quadratic equation √3x² - 2√2x - 2√3 = 0​ are $\sqrt{6}$ and $-\dfrac{\sqrt{2}}{\sqrt{3}}$.

Solution :-

The equation is - √3x² - 2√2x - 2√3 = 0​

Here,

a = √3,

b = - 2√2,

c = - 2√3

Quadratic Equation -

x = $\dfrac{-b + \sqrt{b^{2}-4ac } }{2a}$ or $\dfrac{-b - \sqrt{b^{2}-4ac } }{2a}$

                                                   

$\star$ Finding 1st root -

x = $\dfrac{-(-2\sqrt{2} ) + \sqrt{(-2\sqrt{2})^{2}-(4\times \sqrt{3}\times -2\sqrt{3}} }{2\times \sqrt{3}}$

x = $\dfrac{2\sqrt{2} + \sqrt{8-(-24) }}{2\sqrt{3}}$

x =  $\dfrac{2\sqrt{2} + \sqrt{8+24 }}{2\sqrt{3}}$

x =  $\dfrac{2\sqrt{2} + \sqrt{32}}{2\sqrt{3}}$

x =  $\dfrac{2\sqrt{2} + 4\sqrt{2}}{2\sqrt{3}}$

x =  $\dfrac{6\sqrt{2}}{2\sqrt{3}}$

x =  $\dfrac{3\sqrt{2}}{\sqrt{3}}$

Multiplying the fraction with $\dfrac{\sqrt{3}}{\sqrt{3} }$.

x = $\dfrac{3\sqrt{2}}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}$

x = $\dfrac{3\sqrt{6}}{3}$  

x = $\sqrt{6}$

                                                   

$\star$ Finding 2nd root:-

x = $\dfrac{-(-2\sqrt{2} ) - \sqrt{(-2\sqrt{2})^{2}-(4\times \sqrt{3}\times -2\sqrt{3}} }{2\times \sqrt{3}}$

x =  $\dfrac{2\sqrt{2} - \sqrt{8-(-24) }}{2\sqrt{3}}$

x =  $\dfrac{2\sqrt{2} - \sqrt{8+24 }}{2\sqrt{3}}$

x = $\dfrac{2\sqrt{2} - \sqrt{32}}{2\sqrt{3}}$

x = $\dfrac{2\sqrt{2} - 4\sqrt{2}}{2\sqrt{3}}$

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

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

                                                   

The roots of x in the quadratic equation √3x² - 2√2x - 2√3 = 0​ are $\sqrt{6}$ and $-\dfrac{\sqrt{2}}{\sqrt{3}}$.