Question

write discriminator of √3x^2-2√2x-2√3=0

Answer 1

Answer:

Step-by-step explanation:

Given : \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0    

On comparing with ax^2 + bx + c = 0, we get

= > a = \sqrt{3},b = -2\sqrt{2},c =  -2\sqrt{3}    

(1)

x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}    

= > \frac{-(-2\sqrt{2})+\sqrt{(-2\sqrt{2})^2 - 4\sqrt{3}(-2\sqrt{3})}}{2\sqrt{3}}        

= > \frac{2\sqrt{2}\sqrt{(2\sqrt{2})^2 + 4\sqrt{3} * 2\sqrt{3}}}{2\sqrt{3}}        

= > \frac{2\sqrt{2}+\sqrt{(2\sqrt{2})^2 + 24}}{2\sqrt{3}}      

= > \frac{2\sqrt{2} + \sqrt{32}}{2\sqrt{3}}      

= > \frac{2\sqrt{2} + 4\sqrt{2}}{2\sqrt{3}}      

= > \frac{6\sqrt{2}}{2\sqrt{3}}    

= > \frac{3\sqrt{2}}{\sqrt{3}}    

= > \sqrt{6}  

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(2)

= > x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}    

= > x = \frac{-(-2\sqrt{2})-\sqrt{(2\sqrt{2})^2+4\sqrt{3}*2\sqrt{3}}}{2\sqrt{3}}        

= >\frac{2\sqrt{2}-{4\sqrt{2}}}{2\sqrt{3}}      

= > \frac{-2\sqrt{2}}{2\sqrt{3}}    

= > -\sqrt{\frac{2}{3}}    

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Therefore the required quadratic solutions are|:

= > x = \boxed {\sqrt{6},  -\sqrt{\frac{2}{3}}  }  

Hope this helps

Answer 2

I hope this will help you