Question

Find the roots of the equation: 3x²-6x+2=0

Answer 1

Given: 3x²-6x+2=0

We have to apply the equation:

$\boxed{\sf{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}}$

Here,

• a = 3

• b = -6

• c = 2

Applying values to the equation:

$\sf{\implies x=\dfrac{-(-6)\pm\sqrt{(-6)^2-(4\times3\times2)}}{(2\times3)}}$

$\sf{\implies x=\dfrac{6\pm\sqrt{36-24}}{6}}$

$\sf{\implies x=\dfrac{6\pm\sqrt{12}}{6}}$

$\sf{\implies x=\dfrac{6\pm2\sqrt{3}}{6}}$

$\sf{\implies x=\dfrac{6+2\sqrt{3}}{6} \ \ \ \ \& \ \ \ \ x=\dfrac{6-2\sqrt{3}}{6}}$

$\boxed{\bold{\sf{\therefore x=\dfrac{6+2\sqrt{3}}{6} \ \ \ \ \& \ \ \ \ x=\dfrac{6-2\sqrt{3}}{6}}}}$

Answer 2

given: 3x%C2%B2-6x+2=0We have to apply the equation:Here,a = 3b = -6c = 2Applying values to the equation: