Question

determine the roots of equations √3x2 -2x - √3=0​

Answer 1

Answer:

$\frac{ \sqrt{3} }{2}$

Answer 2

ANSWER:

Given:

• √3x^2 - 2x - √3 = 0

To Find:

• Roots of the equation.

Solution:

$\text{We are given that,}\\\\:\longrightarrow\sqrt3x^2-2x-\sqrt3=0\\\\\text{We know that by Quadratic Formula,}\\\\\text{For a quadratic equation: ax^2+bx+c=0 }\\\\:\hookrightarrow x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{So, for \sqrt3x^2-2x-\sqrt3=0 ,}\\\\:\longrightarrow a=\sqrt3,\: b=-2,\:c=-\sqrt3\\\\\text{Hence,}\\\\:\implies x=\dfrac{-(2)\pm\sqrt{(-2)^2-4(\sqrt3)(-\sqrt3)}}{2(\sqrt3)}\\\\:\implies x=\dfrac{-2\pm\sqrt{4+(4\times3)}}{2\sqrt3}\\\\:\implies x=\dfrac{-2\pm\sqrt{4+12}}{2\sqrt3}$

$:\implies x=\dfrac{-2\pm\sqrt{16}}{2\sqrt3}\\\\:\implies x=\dfrac{-2\pm4}{2\sqrt3}\\\\:\implies x=\dfrac{2\!\!\!/(-1\pm2)}{2\!\!\!/\sqrt3}\\\\:\implies x=\dfrac{-1\pm2}{\sqrt3}\\\\\text{Hence,}\\\\:\implies x=\dfrac{-1+2}{\sqrt3}\:\:\:and\:\:\:x=\dfrac{-1-2}{\sqrt3}\\\\:\implies x=\dfrac{1}{\sqrt3}\:\:\:and\:\:\:x=\dfrac{-3}{\sqrt3}\\\\\bf{:\implies x=\dfrac{1}{\sqrt3}\:\:and\:\:-\sqrt3}$

Formula Used:

$\text{Quadratic Formula}\implies x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$