Question

solve:
$4 \sqrt{3} x^{2} + 5x - 2 \sqrt{3} = 0$
by b²-4ac method.....♪​

Answer 1

We have :-

• a = 4√3
• b = 5
• c = -2√3

We know, for x :

$\sf{\dfrac{-b \pm \sqrt{D} }{2a} }\\\\\displaystyle{\sf{= \frac{-5 \pm \sqrt{(5)^2-4\times 4 \sqrt{3} \times -2 \sqrt{3}} }{2\times4\sqrt{3}} }}\\\\\displaystyle{\sf{= \dfrac{-5 \pm \sqrt{25-(-32 \times 3)} }{8\sqrt{3} } }}\\\\\displaystyle{\sf{= \dfrac{-5 \pm\sqrt{25+96} }{8\sqrt{3} } }}\\\\\displaystyle{\sf{= \dfrac{-5 \pm \sqrt{121} }{8\sqrt{3} } }}\\\\\displaystyle{\sf{= \dfrac{-5 \pm 11}{8\sqrt{3} } }}\\\\\displaystyle{\sf{= \dfrac{-5+11}{8\sqrt{3} } \:\: or \:\: \frac{-5-11}{8\sqrt{3} } }}$

$\displaystyle{\sf{= \dfrac{6}{8\sqrt{3} }\:\: or \:\: \frac{-16}{8\sqrt{3} } }}\\\\\displaystyle{\sf{= \dfrac{3}{4\sqrt{3} }\:\: or \:\: \frac{-2}{\sqrt{3} } }}\\\\\boxed{\displaystyle{\sf{= \dfrac{\sqrt{3} }{4}\:\: or \:\: \frac{-2}{\sqrt{3} } }}}$

Hence, the values of x are √3/4 or -2/√3.

Answer 2

What is a Quadratic Equation?

If ax²+bx+c is a quadratic polynomial,it is equal to zero,then the following mathematical expression is said to be a Quadratic Equation.

•There are three ways to find the roots of an equation:

1. Splitting the middle term

2. Completing the square

3. Quadratic Formula

•Given equation,

4√3x²+5x-2√3=0

On comparison with general form of quadratic equation,ax²+bx+c=0

Here,

a=4√3,b=5 and c= -2√3

Now,

Discriminant,

D=b²-4ac

=(5)²-4(4√3)(-2√3)

=25+96

=121

Applying quadratic formula:

$x = \frac{ - 5± \sqrt{25 + 96} }{8 \sqrt{3} }$

$x = \frac{ - 5±11}{8 \sqrt{3} }$

Hence,

x=√3/4 or x=-2/√3