Question

Find the value of x in the given quadratic equation: $\sqrt{3}{{x}^{2}}-\sqrt{2}x+3\sqrt{3}=0$

Answer 1

Hint: As we know this is a quadratic equation we will use the sridharacharya formula to find the roots of this quadratic equation. After that we will check whether the roots are real or complex, if it is complex then we will convert it in the form of iota(i) and that will be our final answer.

Complete step-by-step solution -
The conjugate of a + ib is a – ib.
First we will use sridharacharya formula to find the roots of this quadratic equation,
The formula is:
If the equation is $a{{x}^{2}}+bx+c=0$ , then the formula for finding x is:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now we will use this formula to find the value of x in $\sqrt{3}{{x}^{2}}-\sqrt{2}x+3\sqrt{3}=0$:
$\begin{align}
  & x=\dfrac{-\left( -\sqrt{2} \right)\pm \sqrt{{{\left( \sqrt{2} \right)}^{2}}-4\times 3\sqrt{3}\times \sqrt{3}}}{2\sqrt{3}} \\
 & x=\dfrac{\sqrt{2}\pm \sqrt{2-4\times 3\times 3}}{2\sqrt{3}} \\
 & x=\dfrac{\sqrt{2}\pm \sqrt{-34}}{2\sqrt{3}} \\
 & x=\dfrac{\sqrt{2}\pm \sqrt{34}i}{2\sqrt{3}} \\
 & x=\dfrac{1\pm \sqrt{17}i}{\sqrt{6}} \\
\end{align}$
Here the value of i is $\sqrt{-1}$ .
Therefore, we have solved the given quadratic equation and have found both the values of x.
Hence both the solutions are complex and conjugate to each other.

Note: Here in this question students might get confused seeing the coefficient of different power of x is irrational and can make a mistake. And thing should also kept in mind that if the coefficients are complex then it is not true that the roots of quadratic equation will in conjugate pairs but if all the coefficients are real then it is always true that if one solution is complex then there must be another and both of them must conjugate pair to each other.