Question

find the quadratic equation 3x^2-4√3x+4​

Answer 1

Solution:

$3x {}^{2} - 4 \sqrt{3} x + 4 = 0 \\ \\ = > 3x {}^{2} - 2 \sqrt{3} x - 2 \sqrt{3} x + 4 = 0 \\ \\ = > \sqrt{3} x( \sqrt{3} x - 2) - 2( \sqrt{ 3 } x - 2) = 0 \\ \\ = > ( \sqrt{3}x - 2)( \sqrt{3} x - 2) = 0 \\ \\ = > x = \frac{2}{ \sqrt{3} }$

This shows that the given quadratic equation has real and equal roots. =>D=0

I have done this equation by splitting the middle term.

Now,

3×4= -2√3x × -2√3x =12

and -2√3x -2√3x= -4√3x

Answer 2

3x²-4√3x+4

3x²-2√3x-2√3x+4

(3x-2√3)(x-√3)