Question

Find the roots of the quadratic equations root 3x^2-2x-root3

Answer 1

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Answer 2

Answer:

The roots are $x=\frac{1+\sqrt{(1+3\sqrt{3}}}{3},\frac{1-\sqrt{(1+3\sqrt{3}}}{3}$

Step-by-step explanation:

Given : The quadratic equation  $y=3x^2-2x-\sqrt{3}$

To find : The roots of the quadratic equation?

Solution :

Solving by discriminant method  

General form - $ax^2+bx+c=0$

$D=b^2-4ac$

Solution is $x=\frac{-b\pm\sqrt{D}}{2a}$

Equation is $3x^2-2x-\sqrt{3}=0$

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

$D=b^2-4ac$

$D=(-2)^2-4(3)(\sqrt{3})$

$D=4(1+3\sqrt{3})$

Solution is $x=\frac{-b\pm\sqrt{D}}{2a}$

$x=\frac{-(-2)\pm\sqrt{4(1+3\sqrt{3})}}{2(3)}$

$x=\frac{2\pm2\sqrt{(1+3\sqrt{3}}}{6}$

$x=\frac{1\pm\sqrt{(1+3\sqrt{3}}}{3}$

Therefore, The roots are $x=\frac{1+\sqrt{(1+3\sqrt{3}}}{3},\frac{1-\sqrt{(1+3\sqrt{3}}}{3}$