Question

3x²-4√3x+4=0
$3x$
​

Answer 1

Step-by-step explanation:

hope it helps you

follow me....

Answer 2

$3x² - 4\sqrt{3x} + 4 = 0$

Comparing it with $ax² + bx + c = 0$ , we get

a = 3 , b = - 4$\sqrt{3}$ & c = 4

Discriminant $= b² - 4ac$

$= (-4 \sqrt{3})² - 4(3)(4)$

$= 48 - 48 = 0$

Since $b² - 4ac = 0$

Real roots exist for the given equation and they are equal to each other . The roots will be

$\frac {-b}{2a}$ and $\frac {-b}{2a}$

$\implies \frac {-b}{2a} = - \frac {(-4\sqrt{3})}{2 \times 3} = \frac {4\sqrt{3}}{6} = \frac {2\sqrt{3}}{3} = \frac {2}{\sqrt3}$

Therefore , the roots are $\frac {2}{\sqrt3}$ and $\frac {2}{\sqrt{3}}$