Math, asked by sumitkushwahak2005, 2 months ago

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

Answers

Answered by yashanbali
0

Answer:

the answer of this question is 0

Answered by MrImpeccable
12

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}

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