Question

find the roots of the quadratic equation 2 root 3x^2-5x+root3=0 if they exist by using quadratic formula

Answer 1

2√3x² - 5x + √3 = 0
2√3x² - 3x - 2x + √3 = 0
√3x(2x - √3) - 1(2x - √3) = 0
(2x - √3)(√3x - 1) = 0
x = √3/2, 1/√3

Answer 2

Answer:

$\text{The roots are }\frac{3}{2\sqrt3}, \frac{1}{\sqrt3}$

Step-by-step explanation:

Given the equation

$2\sqrt3 x^2-5x+\sqrt3=0$

we have to find the roots of above equation using quadratic formula.

$2\sqrt3 x^2-5x+\sqrt3=0$

$\text{Comparing above equation with }ax^2+bx+c=0\text{ , we get}$

$a=2\sqrt3, b=-5, c=\sqrt3$

By quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-5)\pm \sqrt{(-5)^2-4(2\sqrt3)(\sqrt3)}}{2(2\sqrt3)}$

$x=\frac{5\pm\sqrt{25-24}}{4\sqrt3}$

$x=\frac{5\pm 1}{4\sqrt3}$

$x=\frac{6}{4\sqrt3}, \frac{4}{4\sqrt3}$

$\text{Hence, the roots are }\frac{3}{2\sqrt3}, \frac{1}{\sqrt3}$