Question

Solve $3{{x}^{2}}-5x+2=0$ by completing the square method.

Answer 1

Hint: We will first start by dividing the whole equation by 3. Then we will add and subtract the square of the coefficient of x divided by 2 in the given equation and then use the formula that ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a+b \right)}^{2}}$and then further solve the equation for value of x.

Complete step-by-step answer:
Now, we have the equation given to us as $3{{x}^{2}}-5x+2=0$.
Now, we will divide the equation with the value of coefficient of ${{x}^{2}}$ that is 3. So, we have the equation as ${{x}^{2}}-\dfrac{5}{3}x+\dfrac{2}{3}=0$.
Now, we will add and subtract the square of the coefficient of x divided by 2 in the given equation. So, we have,
$\begin{align}
  & {{x}^{2}}-\dfrac{5}{3}x+{{\left( \dfrac{5}{6} \right)}^{2}}-{{\left( \dfrac{5}{6} \right)}^{2}}+\dfrac{2}{3}=0 \\
 & {{x}^{2}}-\dfrac{5}{3}x+{{\left( \dfrac{5}{6} \right)}^{2}}={{\left( \dfrac{5}{6} \right)}^{2}}-\dfrac{2}{3} \\
\end{align}$
Now, we know that the value of expression ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$. So, using this we have,
$\begin{align}
  & {{\left( x-\dfrac{5}{6} \right)}^{2}}=\dfrac{25-24}{36} \\
 & {{\left( x-\dfrac{5}{6} \right)}^{2}}=\dfrac{1}{36} \\
 & \left( x-\dfrac{5}{6} \right)=\pm \sqrt{\dfrac{1}{36}} \\
 & \left( x-\dfrac{5}{6} \right)=\pm \dfrac{1}{6} \\
 & x=\dfrac{5}{6}\pm \dfrac{1}{6} \\
\end{align}$
So, we have two values of x as,
$\begin{align}
  & x=\dfrac{5+1}{6}\ or\ x=\dfrac{5-1}{6} \\
 & x=\dfrac{6}{6}\ or\ x=\dfrac{4}{6} \\
 & x=1\ or\ x=\dfrac{2}{3} \\
\end{align}$
Therefore, the value of x is $1,\dfrac{2}{3}$.

Note: It is important to note that we have used a fact that if ${{x}^{2}}=a$ then $x=\pm \sqrt{a}$. So, we have two values of x one negative and other positive. It is important to add the $\pm $ sign, so that one solution of the equation is not missed. It is a very common mistake that should be kept in mind.