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

Last updated date: 22nd Jun 2024
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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, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

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, $x = \dfrac{2}{{\sqrt 3 }}$.