Question

Find the roots of the quadratic equation \[{\text{3}}{{\text{x}}^2}{\text{ - 4}}\sqrt 3 {\text{x + 4 = 0}}\]

Answer 1

Hint: We will use formula of Sridhar acharya here. Which is given as, if, a quadratic equation, \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] then, the roots of the equations would be, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step answer:

Here, the given equation for us is, \[{\text{3}}{{\text{x}}^2}{\text{ - 4}}\sqrt 3 {\text{x + 4 = 0}}\].
The above equation is in the form of \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\]
$\Rightarrow$ \[a = 3,b = - 4\sqrt 3 ,c = 4\]
So, now if we use the formula of Sridhar acharya, we will get,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
On substituting values of a, b and c, we get,
\[ \Rightarrow x = \dfrac{{ - ( - 4\sqrt 3 ) \pm \sqrt {{{( - 4\sqrt 3 )}^2} - 4.3.4} }}{{2.3}}\]
\[ \Rightarrow x = \dfrac{{4\sqrt 3 \pm \sqrt {16 \times 3 - 48} }}{6}\]
\[ \Rightarrow x = \dfrac{{4\sqrt 3 \pm \sqrt {48 - 48} }}{6}\]
\[ \Rightarrow x = \dfrac{{4\sqrt 3 }}{6}\]
\[ \Rightarrow x = \dfrac{{2\sqrt 3 }}{3}\]
\[ \Rightarrow x = \dfrac{2}{{\sqrt 3 }}\]
So, we have the solution of the equation, \[{\text{3}}{{\text{x}}^2}{\text{ - 4}}\sqrt 3 {\text{x + 4 = 0}}\] as \[x = \dfrac{2}{{\sqrt 3 }}\].

Note: Here in this equation, we get our discriminant, \[\sqrt {{b^2} - 4ac} \]\[ = 0\] which means this quadratic equation will have a equal root. That means the two roots of the equation will be equal and of the same sign. We can also approach this problem in a different way. We can write this equation as \[{(\sqrt 3 x - 2)^2}\]and proceed with the problem to find, \[x = \dfrac{2}{{\sqrt 3 }}\].