Question

Solve the following quadratic equation for x: 4√3x² +5x-2√3 = 0
​

Answer 1

Step-by-step explanation:

43x2+5x−23=0

D=b2−4ac

   =(5)2−4×43×−23

   =25+96

   =121>0

So , roots of given equation are real and distinct 

Answer 2

ANSWER:

To Solve:

• 4√3x² + 5x - 2√3 = 0

Solution:

We are given that,

⇒ 4√3x² + 5x - 2√3 = 0

We know that, for a quadratic equation, ax² + bx + c = 0,

Quadratic Equation says:

$\sf\implies x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

So, in this case, a = 4√3, b = 5, c = -2√3.

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

So,

$\sf\implies x=\dfrac{-(5\pm\sqrt{25+96}}{8\sqrt3}$

$\sf\implies x=\dfrac{-(5\pm\sqrt{121}}{8\sqrt3}$

$\sf\implies x=\dfrac{-(5\pm11}{8\sqrt3}$

Hence,

⇒ x = (-5+11)/8√3 and x = (-5-11)/8√3

So,

⇒ x = 6/8√3 and x = -16/8√3

⇒ x = 3/4√3 and x = -2/√3

On simplifying,

⇒ x = √3/4 and -2√3/3