Question

the roots of 3x square-2x +1÷3=0​

Answer 1

Step-by-step explanation:

3x^2-2x+1=0×3

3x^2-2x+1=0

a=3 b=-2 c=1

x=-b+-√b^2-4ac

Answer 2

Answer:

$3 {x}^{2} - 2x + \frac{1}{3} = 0 \\ (\sqrt{3} {x)}^{2} - 2x +( \frac{1}{ \sqrt{3} } ) {}^{2} = 0 \\ ( \sqrt{3} {x)}^{2} - 2 x\sqrt{3} \times \frac{1}{ \sqrt{3} } + (\frac{1}{ \sqrt{3} } ) {}^{2} \\ it \: is \: of \: the \: for \: (a - b) {}^{2} = {a}^{2} - 2ab + {b}^{2} ) \\ so \\ ( \sqrt{3} x - \frac{1}{ \sqrt{3} } ) {}^{2} = 0 \\ \\ ( \sqrt{3} x - \frac{1}{ \sqrt{3} } ) = 0 \\ \sqrt{3x} = \frac{1}{ \sqrt{3} } \\ x = \frac{1}{3}$