Question

Solve following equation
$\sqrt{3}{{x}^{2}}-2\sqrt{3}x+3\sqrt{3}=0$

Answer 1

If we have a polynomial of degree ‘n’ then we have ‘n’ number of roots or factors. A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula, etc. The quadratic formula is used when we fail to find the factors of the equation.

Complete answer:
Given, $\sqrt{3}{{x}^{2}}-2\sqrt{3}x+3\sqrt{3}=0$.
On comparing the given equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0$. We get, $a=\sqrt{3}$, $b=-2\sqrt{3}$ and $c=3\sqrt{3}$.
Substituting in the formula of standard quadratic, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
$\Rightarrow x=\dfrac{-\left( -2\sqrt{3} \right)\pm \sqrt{{{\left( -2\sqrt{3} \right)}^{2}}-4\left( \sqrt{3} \right)\left( 3\sqrt{3} \right)}}{2\left( \sqrt{3} \right)}$
$\Rightarrow x=\dfrac{2\sqrt{3}\pm \sqrt{\left( 4\times 3 \right)-4\left( 3\times 3 \right)}}{2\sqrt{3}}$
$\Rightarrow x=\dfrac{2\sqrt{3}\pm \sqrt{\left( 12 \right)-4\left( 9 \right)}}{2\sqrt{3}}$
$\Rightarrow x=\dfrac{2\sqrt{3}\pm \sqrt{12-36}}{2\sqrt{3}}$
$\Rightarrow x=\dfrac{2\sqrt{3}\pm \sqrt{-24}}{2\sqrt{3}}$
$\Rightarrow x=\dfrac{2\sqrt{3}\pm \sqrt{-6\times 4}}{2\sqrt{3}}$
We know that 4 is a perfect square,
$\Rightarrow x=\dfrac{2\sqrt{3}\pm 2\sqrt{-6}}{2\sqrt{3}}$
$\Rightarrow x=\dfrac{\sqrt{3}\pm \sqrt{-6}}{\sqrt{3}}$
We know that $\sqrt{-1}=i$, then it becomes
$\Rightarrow x=\dfrac{\sqrt{3}\pm i\sqrt{6}}{\sqrt{3}}$
Thus, we have two roots,
$\Rightarrow x=\dfrac{\sqrt{3}+i\sqrt{6}}{\sqrt{3}}$ and $\Rightarrow x=\dfrac{\sqrt{3}-i\sqrt{6}}{\sqrt{3}}$. This is the required result.

Note:
On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. We cannot solve this by simple factorization. That is by expanding the middle term into a sum of two terms, such that the product of two terms is equal to the product of ‘a’ and ‘c’, the sum of two terms is equal to ‘b’. For factorization, the standard equation is rewritten as $a{{x}^{2}}+{{b}_{1}}x+{{b}_{2}}x+c=0$ such that${{b}_{1}}\times {{b}_{2}}=ac$ and${{b}_{1}}+{{b}_{2}}=b$.