Question

Solve the following quadratic using Sridhar Acharya formula: $\sqrt{3}{{x}^{2}}-\sqrt{2}x+3\sqrt{3}=0$

Answer 1

Hint: We first try to explain the formula of Sridhar Acharya. We express the roots of the general equation of quadratic $a{{x}^{2}}+bx+c=0$. We find the coefficients of the given quadratic and place them in the equation. We get two root values of the quadratic.

Complete step-by-step solution:
We first explain the Sridhar Acharya formula. We use the formula to find the roots of a quadratic equation.
For a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
We have been given a quadratic equation $\sqrt{3}{{x}^{2}}-\sqrt{2}x+3\sqrt{3}=0$.
We try to equate it with the general form of the quadratic equation and get
$a=\sqrt{3},b=-\sqrt{2},c=3\sqrt{3}$. We place those values of a, b, c in the equation of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and get
$x=\dfrac{\sqrt{2}\pm \sqrt{{{\left( -\sqrt{2} \right)}^{2}}-4\times \sqrt{3}\times 3\sqrt{3}}}{2\sqrt{3}}=\dfrac{\sqrt{2}\pm \sqrt{-34}}{2\sqrt{3}}$.
Here we get imaginary roots. We have negative value inside the root. So, we take $\sqrt{-1}=i$.
The roots become $x=\dfrac{\sqrt{2}\pm i\sqrt{34}}{2\sqrt{3}}=\dfrac{1\pm i\sqrt{17}}{\sqrt{6}}$.
We have two roots for the quadratic equation.

Note: The root part in the formula of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the determinant. This is a very important part to find the characteristics of the roots. If ${{b}^{2}}-4ac>0$, then the roots are real and unequal. If ${{b}^{2}}-4ac=0$, then the roots are real and equal. If ${{b}^{2}}-4ac<0$, then the roots are imaginary. For our given equation $\sqrt{3}{{x}^{2}}-\sqrt{2}x+3\sqrt{3}=0$, the determinant value was negative and that’s why we got imaginary value.