Question

3x²-2x-1=0 solve by the the method of completion of square

Answer 1

Answer:

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

$3x { }^{2} - (3- 1)x - 1 = 0$

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

$3x(x - 1) + (x - 1) = 0$

$(3x + 1)(x - 1) = 0$

$3x + 1 = 0 \\ x = - \frac{1}{3}$

$x - 1 = 0 \\ x = 1$

Answer 2

Answer:

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

$⇒3 {x}^{2} + 2x = 1$

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

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

$⇒ \: \: \: \: \: {(x + \frac{1}{3} )}^{2} \: = \frac{1}{9} + \frac{1}{3} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1 + 3}{9}$

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

$On \: taking \: root, \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x + \frac{1}{3} = ± \frac{2}{3}$

$On \: taking ( + ) \: sign, \\ \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{2}{3} - \frac{1}{3} = \frac{2 - 1}{3} = \frac{1}{3}$

$On \: taking \: ( - ) \: sign, \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: x + \frac{1}{3} = \frac{ - 2}{3}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{ - 2}{3} - \frac{1}{3}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{ - 3}{3} = - 1$

$∴ Roots \: of \: equation \: are \: x = \frac{1}{3} , - 1.$