Question

Find the roots of the following quadratic equations if they exist, by the method of square: $4{{x}^{2}}+4\sqrt{3}x+3=0$.

Answer 1

Hint: To solve this question, we should know the concept of completing the square method. In this method, we have to represent a quadratic equation of the type $a{{x}^{2}}+bx+c=0$ in the form $a{{\left( x+d \right)}^{2}}+c=0$.

Complete step-by-step answer:
In this question, we are given a quadratic equation of the type $a{{x}^{2}}+bx+c=0$ and we have to convert it in the form of $a{{\left( x+d \right)}^{2}}+c=0$. So, we have been given the quadratic equation as $4{{x}^{2}}+4\sqrt{3}x+3=0$. To find the roots of any quadratic equation by the method of square, we have to follow a few steps.
We have been given, $4{{x}^{2}}+4\sqrt{3}x+3=0$. Now, we will take 4 common from the left hand side of the equation, so we get, $4\left( {{x}^{2}}+\sqrt{3}x+\dfrac{3}{4} \right)=0$.
$\Rightarrow {{x}^{2}}+\sqrt{3}x+\dfrac{3}{4}=0$
Now, on shifting $\dfrac{3}{4}$ to the right side, we get,
${{x}^{2}}+\sqrt{3}x=-\dfrac{3}{4}\ldots \ldots \ldots (i)$
Now, we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\Rightarrow {{\left( x+a \right)}^{2}}={{x}^{2}}+{{a}^{2}}+2ax\ldots \ldots \ldots (ii)$
Now we have to find the value of $a$ in the above equation, so that we can represent equation (i) in the form of equation (ii). Now, if we look at the left hand side of equation (i) and compare it with the right hand side of equation (ii), we will observe that terms $\sqrt{3}x$ and $2ax$ are same as they are the coefficients of $x$ and can be equated to each other as below,
$\begin{align}
  & 2a=\sqrt{3} \\
 & \Rightarrow a=\dfrac{\sqrt{3}}{2} \\
\end{align}$
To complete the square, we will add ${{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}$ on both sides of the equation, so we get,
${{x}^{2}}+2\left( \dfrac{\sqrt{3}}{2} \right)x+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}=-\dfrac{3}{4}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}$
${{x}^{2}}+2\left( \dfrac{\sqrt{3}}{2} \right)x+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}=-\dfrac{3}{4}+\dfrac{3}{4}$
${{x}^{2}}+2\left( \dfrac{\sqrt{3}}{4} \right)x+{{\left( \dfrac{\sqrt{3}}{4} \right)}^{2}}=0\ldots \ldots \ldots (iii)$
 From the equations (ii) and (iii), we get, ${{\left( x+\dfrac{\sqrt{3}}{2} \right)}^{2}}=0$
$\begin{align}
  & \Rightarrow x+\dfrac{\sqrt{3}}{2}=0 \\
 & \Rightarrow x=-\dfrac{\sqrt{3}}{2} \\
\end{align}$
Therefore, the roots of the quadratic equation $4{{x}^{2}}+4\sqrt{3}x+3=0$ are real and equal, that is, $x=-\dfrac{\sqrt{3}}{2}$ is the root of the quadratic equation.

Note: To check the existence of roots, we can use a discriminant formula, that is, $D={{b}^{2}}-4ac$, if $D>0$ or $D=0$, then the roots exist.