Question

find the nature of roots of quadratic equation 3 x square minus 4 root 3 X + 4 is equal to zero

Answer 1

I hope this answer would be helpful for you

Answer 2

Answer:

Nature of roots = Roots are real and equal.

Roots are  $x=\frac{-2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$

Step-by-step explanation:

Given : $3x^2-4\sqrt3x +4=0$

To find : Nature of roots of quadratic equation

Solution : We use Discriminant method to solve the quadratic equation.

Quadratic equation form   $ax^2+bx+c=0$

$D=b^2-4ac$

If D>0 , roots are real and distinct

If D=0 , roots are real and equal

If D<0 , roots are complex and distinct

So, we find the D of equation  $3x^2-4\sqrt3x +4=0$

where a= 3 , b= 4√3 , c=4

$D=(4\sqrt3)^2-4(3)(4)$

$D=16(3)-48$

$D=48-48=0$

Which implies D=0 , Roots are real and equal.

Solution to find roots → $x=\frac{-b\pm\sqrt{D}}{2a}$

Put values,

$x=\frac{-4\sqrt{3}\pm\sqrt{0}}{2(3)}$

$x=\frac{-4\sqrt{3}}{6}$

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

Roots are $x=\frac{-2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$