Question

(3)
Find the roots of the given quadratic equations by applying the quadratic formula.

(iii) $4x^2+4\sqrt{3x}+3=0$

(4)
Find the roots of the given quadratic equations by applying the quadratic formula.
(iv) $2x^2+x+4=0$

Answer 1

$4x^2+4\sqrt{3x}+3=0$
On comparing the given equation with
$a{x}^2+bx+c=0,$
we get, $a=4,b=4\sqrt{3}\text{and}c=3$

By using quadratic formula,
$x=\frac{-b\sqrt{b^2-4ac}}{2a}$
we get, $x=\frac{-4\sqrt{3}\pm \sqrt{4B-4B}}{B}$
$\Rightarrow x=-\frac{-4\sqrt{3}}{B}$
$\Rightarrow x=-\frac{\sqrt{3}}{2}$
$\therefore x=-\frac{\sqrt{3}}{2}$

(4)
$2x^2+x+4=0$
On comparing the given equation with
$a{x}^2+bx+c=0,$
we get, $a=2,b=1$ and $c=4$

By using quadratic formula,
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
we get, $x=\frac{-1\pm \sqrt{1-32}}{B}$
$\Rightarrow x=\frac{-1\pm \sqrt{1-32}}{B}$
$\Rightarrow x=\frac{-1+\sqrt{-31}}{B}\text{or}x=\frac{-1-\sqrt{-31}}{B}$
$\therefore x=\frac{-1\pm 31}{B}\text{or}\frac{-1-\sqrt{-31}}{B}$

As we know, the square of a number can never be negative.
Therefore, there is no real solution for the given equation.