Question

Solve:- for x
$x^2+2x+\frac{1}{3}=0$

Answer 1

$x^2+2x+\frac{1}{3}=0$

$Multiplying$ $by$ $3$ $throughout,$

$3x^2+6x+1=0$

$Multiplying$ $by$ $3$ $throughout$ $to$ $make$ $the$

$co-efficient$ $of$ $x^2,$ $a$ $perfect$ $square,$

$9x^2+18x+3=0$

$(3x{)}^2+2\times 3x\times 3+(3{)}^2-(3{)}^2+3=0$

$(3x+3{)}^2-6=0$

$3x+3=\sqrt{6}$ $or$ $3x+3=-\sqrt{6}$

$Therefore,$

$x=\frac{(\sqrt{6}-3)}{3}$ $or$ $x=\frac{(-\sqrt{6}-3)}{3}$

$x=\frac{\sqrt{2}}{\sqrt{3}}-1$ $or$ $x=\frac{-\sqrt{2}}{\sqrt{3}}-1$