Question

solve the equation by completing squaring method 3x²-4x+4​

Answer 1

Your Question is:

$3 {x}^{2} + 4x - 4 = 0$

Solution:

$3 {x}^{2} + 4x - 4 = 0$

$3 {x}^{2} + (6 - 2)x - 4 = 0$

$3 {x}^{2} + 6x - 2x - 4 = 0$

$3x(x + 2) - 2(x + 2) = 0$

$(x + 2)(3x - 2) = 0 \\ x = - 2 \: \: or \: x = \frac{3}{2}$

Verification,

Substituting the value of x=3 in the give equation,

$3 {x}^{2} + 4x - 4 = 0$

$3( { - 2}^{2} ) + 4( - 2) - 4 = 0 \\ = > 3(4) - 8 - 4 = 0 \\ = > 12 - 12 = 0 \\ = > 0 = 0$

Again Now substituting the another value of x which is x=2/3 in the same given equation we get,

$3 {x}^{2} + 4x - 4 = 0 \\3 { \frac{2}{3} }^{2} + 4 \frac{2}{3} - 4 = 0 \\ = > 3 \frac{4}{9} + \frac{8}{3} - 4 = 0$

Now taking LCM of 9 and 3 we get 3 as LCM,

$= > \frac{12}{9} + \frac{8}{3} - 4 = 0 \\ = > \frac{12 + 24 - 36}{9} = 0 \\ = > \frac{36 - 36}{9} = 0 \\ = > \frac{0}{9} = 0 \\ = > 0 = 0$

Therefore the values of x are

$x = - 2 \: and \: x = \frac{2}{3}$