Question

Find the roots of the following quadratic equation:
$4{x^2} + 4\sqrt 3 x + 3 = 0$

Answer 1

Hint: The question is related to the quadratic equation whose highest power is 2. You must know that there are two possible roots of the quadratic equation which have the highest power two. Divide the equation by number 4 to simplify the equation as 4 is common from the first two terms of the quadratic equation. Write the term separately for better understanding of the question.

Complete step-by-step answer:
Given quadratic equation
$4{x^2} + 4\sqrt 3 x + 3 = 0$
Divide the whole equation by 4
${x^2} + \sqrt 3 x + \dfrac{3}{4} = 0$
Now convert the above equation in the formula of
${a^2} + 2ab + {b^2} = {(a + b)^2}$
Here $a = 1,b = \dfrac{{\sqrt 3 }}{2}$
Write the term in separately
${x^2} + 2 \times 1 \times \dfrac{{\sqrt 3 }}{2} + {(\dfrac{{\sqrt 3 }}{2})^2} = 0$
Convert the equation in the formula
$ \Rightarrow {(x + \dfrac{{\sqrt 3 }}{2})^2} = 0$
Taking the root both the sides
$ \Rightarrow (x + \dfrac{{\sqrt 3 }}{2}) = 0$
Equal to 0
\[ \Rightarrow x + \dfrac{{\sqrt 3 }}{2} = 0\]
Subtract $\dfrac{{\sqrt 3 }}{2}$ from both the left-hand side and right- hand side
$\therefore x = - \dfrac{{\sqrt 3 }}{2}$

Hence the root of the quadratic equation is $ - \dfrac{{\sqrt 3 }}{2}$

Note: Use the formula of ${a^2} + 2ab + {b^2} = {(a + b)^2}$ to simplify the question. In mathematics, a quartic equation is one which can be expressed as a quartic function equaling zero. Since a quadratic function is defined by a polynomial of even degree, it has the same infinite limit when the argument goes to positive or negative infinity. If a is positive, then the function increases to positive infinity at the both ends and thus the function has a global minimum. Likewise, if a is negative, it decreases to negative infinity and has a global maximum. In both cases it may or may not have another local maximum and another local minimum.