Question

How do you solve $3{{x}^{2}}-2x-4=0$?

Answer 1

Hint: For this problem they have asked to calculate the solution of the given equation. We can observe that the given equation is a quadratic equation which is in the form of $a{{x}^{2}}+bx+c=0$. For this problem we are going to use the method of completing squares to solve the equation. We know that in method of completing squares we will convert the given equation which is form of $a{{x}^{2}}+bx+c=0$ into $a{{\left( x+d \right)}^{2}}+e=0$ by applying some arithmetic operations. After that we will simplify the obtained equation to get the value of $x$ by taking square root function on both sides of the equation. Now we will simplify the equation to get the required result.

Complete step by step answer:
Given equation, $3{{x}^{2}}-2x-4=0$
We can clearly see that the above equation is a quadratic equation, so we can use the method of completing squares to solve the given equation.
Comparing the given equation with $a{{x}^{2}}+bx+c=0$, then we will get
$a=3$, $b=-2$, $c=-4$.
Dividing the given equation with $3$, then we will get
$\begin{align}
  & \dfrac{3{{x}^{2}}-2x-4}{3}=\dfrac{0}{3} \\
 & \Rightarrow \dfrac{3}{3}{{x}^{2}}-\dfrac{2}{3}x-\dfrac{4}{3}=0 \\
 & \Rightarrow {{x}^{2}}-\dfrac{2}{3}x-\dfrac{4}{3}=0 \\
\end{align}$
In the above equation the coefficient of $x$ is $\dfrac{2}{3}$. Adding the square of half of the coefficient of $x$ in the above equation, then we will get
$\begin{align}
  & {{x}^{2}}-\dfrac{2}{3}x-\dfrac{4}{3}+{{\left( \dfrac{1}{2}.\dfrac{2}{3} \right)}^{2}}-{{\left( \dfrac{1}{2}.\dfrac{2}{3} \right)}^{2}}=0 \\
 & \Rightarrow {{x}^{2}}-\dfrac{2}{3}x-\dfrac{4}{3}+{{\left( \dfrac{1}{3} \right)}^{2}}-{{\left( \dfrac{1}{3} \right)}^{2}}=0 \\
\end{align}$
Rearranging and rewriting the terms in the above equation, then we will get
$\Rightarrow {{x}^{2}}-2\left( \dfrac{1}{3} \right)\left( x \right)+{{\left( \dfrac{1}{3} \right)}^{2}}-\dfrac{4}{3}-\dfrac{1}{9}=0$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, then we will get
$\Rightarrow {{\left( x-\dfrac{1}{3} \right)}^{2}}-\dfrac{4}{3}-\dfrac{1}{9}=0$
Taking the constants in the above equation to the other side of the equation, then we will get
$\begin{align}
  & \Rightarrow {{\left( x-\dfrac{1}{3} \right)}^{2}}=\dfrac{4}{3}+\dfrac{1}{9} \\
 & \Rightarrow {{\left( x-\dfrac{1}{3} \right)}^{2}}=\dfrac{3\times 4+1}{9} \\
 & \Rightarrow {{\left( x-\dfrac{1}{3} \right)}^{2}}=\dfrac{13}{9} \\
\end{align}$
Applying square root on both side of the above equation, then we will get
$\begin{align}
  & \Rightarrow \sqrt{{{\left( x-\dfrac{1}{3} \right)}^{2}}}=\sqrt{\dfrac{13}{9}} \\
 & \Rightarrow x-\dfrac{1}{3}=\pm \dfrac{\sqrt{13}}{3} \\
 & \Rightarrow x=\dfrac{1}{3}\pm \dfrac{\sqrt{13}}{3} \\
\end{align}$

$\therefore $ The solution for the given equation is $x=\dfrac{1\pm \sqrt{13}}{3}$.

Note: We can also check whether the obtained solution is correct or wrong by substituting either $x=\dfrac{1+\sqrt{13}}{3}$ or $x=\dfrac{1-\sqrt{13}}{3}$ in the given equation. Substituting $x=\dfrac{1+\sqrt{13}}{3}$in the given equation $3{{x}^{2}}-2x-4=0$, then we will get
$\begin{align}
  & \Rightarrow 3{{\left( \dfrac{1+\sqrt{13}}{3} \right)}^{2}}-2\left( \dfrac{1+\sqrt{13}}{3} \right)-4=0 \\
 & \Rightarrow 3\left( \dfrac{{{\left( 1+\sqrt{13} \right)}^{2}}}{9} \right)-\dfrac{2+2\sqrt{13}}{3}-4=0 \\
\end{align}$
Simplifying the above equation by using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, then we will get
$\Rightarrow \dfrac{1+{{\left( \sqrt{13} \right)}^{2}}+2\left( 1 \right)\left( \sqrt{13} \right)}{3}-\dfrac{2+2\sqrt{13}}{3}-4=0$
Taking LCM for the fractions we have in the above equation, then we will get
$\begin{align}
  & \Rightarrow \dfrac{1+13+2\sqrt{13}-2-2\sqrt{13}}{3}-4=0 \\
 & \Rightarrow \dfrac{12}{3}-4=0 \\
 & \Rightarrow 4-4=0 \\
 & \Rightarrow 0=0 \\
 & \Rightarrow LHS=RHS \\
\end{align}$
Hence the obtained solution is correct.