Question

solve using perfect square method 3x^2+2x-1=0​

Answer 1

Answer :-

$3 {x}^{2} + 2x - 1 = 0$

Add 1 to both sides of equation.

$3 {x}^{2} + 2x = 1$

Divide each term by 3.

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

$Here,\: a = 1; \: b = \frac{2}{3}; \: c = \frac{1}{3}$

To find a value that is equal to the square of half of b, the coefficient of x.

$({ \frac{b}{2} })^{2} = ({ - \frac { 1}{3} })^{2}$

Add the term to both sides of equation.

${x}^{2} + \frac{2x}{3} + ( { - \frac{1}{3} })^{2} = ( \frac{1}{3} ) + ( { - \frac{1}{3} })^{2}$

After simplifying the equation.

${x}^{2} - \frac{2x}{3} + \frac{1}{9} = \frac{4}{9}$

Here, we can see the expandation of square on LHS.

$( { x - \frac{1}{3} })^{2} = \frac{4}{9}$

$x - \frac{1}{3} = ± \sqrt{ \frac{4}{9} }$

$x - \frac{1}{3} = ±\frac{2}{3}$

$\huge\bold{x = -\frac{1}{3} , 1 }$