Question

do completing of square of quadratic equation 3x^2 - 4√3x + 4 = 0


please answer correctly !!​

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm :\longmapsto\: {3x}^{2} - 4 \sqrt{3}x + 4 = 0$

can be rewritten as

$\rm :\longmapsto\: {3x}^{2} - 4 \sqrt{3}x = - \: 4$

Step :- 1 To make the coefficient of x² unity, divide both sides by 3, we get .

$\rm :\longmapsto\: {x}^{2} - \dfrac{4}{3} \sqrt{3}x = - \: \dfrac{4}{3}$

Step :- 2 Add and subtract the square of half the coefficient of x, we get

$\rm :\longmapsto\: {x}^{2} - \dfrac{4}{3} \sqrt{3}x + \dfrac{4}{3} = - \: \dfrac{4}{3} + \dfrac{4}{3}$

can be rewritten as

$\rm :\longmapsto\: {x}^{2} - \dfrac{4}{3} \sqrt{3}x + \dfrac{4}{3} = 0$

$\rm :\longmapsto\: {x}^{2} - \dfrac{4}{3} \sqrt{3}x + \dfrac{4}{9} \times 3 = 0$

$\rm :\longmapsto\: {x}^{2} - 2 \times \dfrac{2}{3} \sqrt{3} \times x + { \bigg(\dfrac{2 \sqrt{3} }{3} \bigg)}^{2} = 0$

We know,

$\boxed{ \sf{ \: {x}^{2} - 2xy + {y}^{2} = {(x - y)}^{2}}}$

So, using this identity, we get

$\rm :\longmapsto\: { \bigg(x - \dfrac{2 \sqrt{3} }{3} \bigg)}^{2} = 0$

$\rm :\longmapsto\: x - \dfrac{2 \sqrt{3} }{3} = 0$

$\bf :\longmapsto\: x = \dfrac{2 \sqrt{3} }{3}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac