Question

How do you solve ${{\left( x+\dfrac{1}{3} \right)}^{2}}-\dfrac{4}{9}=0$?

Answer 1

Hint: First we will simplify the power of the parenthesis by using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Then we will solve the obtained equation by using a suitable method. We will solve the obtained equation by using the quadratic formula.

Complete step by step solution:
We have been given an equation ${{\left( x+\dfrac{1}{3} \right)}^{2}}-\dfrac{4}{9}=0$
We have to solve the given equation.
Now, we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Now, applying the formula in the given equation we will get
$\Rightarrow {{x}^{2}}+{{\left( \dfrac{1}{3} \right)}^{2}}+2\times x\times \dfrac{1}{3}-\dfrac{4}{9}=0$
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow {{x}^{2}}+\dfrac{1}{9}+\dfrac{2x}{3}-\dfrac{4}{9}=0 \\
 & \Rightarrow {{x}^{2}}-\dfrac{3}{9}+\dfrac{2x}{3}=0 \\
 & \Rightarrow {{x}^{2}}+\dfrac{2x}{3}-\dfrac{1}{3}=0 \\
\end{align}\]
Now, we can solve the given equation by using a quadratic formula. Then we will get
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, on comparing the obtained equation with the general equation $a{{x}^{2}}+bx+c=0$we will get the values as
$a=1,b=\dfrac{2}{3},c=\dfrac{-1}{3}$
Now, substituting the values we will get
$\Rightarrow x=\dfrac{-\left( \dfrac{2}{3} \right)\pm \sqrt{{{\left( \dfrac{2}{3} \right)}^{2}}-4\times 1\times \dfrac{-1}{3}}}{2\times 1}$
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow x=\dfrac{-\left( \dfrac{2}{3} \right)\pm \sqrt{\dfrac{4}{9}+\dfrac{4}{3}}}{2} \\
 & \Rightarrow x=\dfrac{-\left( \dfrac{2}{3} \right)\pm \sqrt{\dfrac{4+12}{9}}}{2} \\
 & \Rightarrow x=\dfrac{-\left( \dfrac{2}{3} \right)\pm \sqrt{\dfrac{16}{9}}}{2} \\
 & \Rightarrow x=\dfrac{-\left( \dfrac{2}{3} \right)\pm \dfrac{4}{3}}{2} \\
\end{align}\]
Now, we have to consider both signs one by one then we will get
$\Rightarrow x=\dfrac{-\left( \dfrac{2}{3} \right)+\dfrac{4}{3}}{2},x=\dfrac{-\left( \dfrac{2}{3} \right)-\dfrac{4}{3}}{2}$
Now, simplifying the obtained equations we will get
$\begin{align}
  & \Rightarrow x=\dfrac{\dfrac{-2+4}{3}}{2},x=\dfrac{\dfrac{-2-4}{3}}{2} \\
 & \Rightarrow x=\dfrac{\dfrac{2}{3}}{2},x=\dfrac{\dfrac{-6}{3}}{2} \\
 & \Rightarrow x=\dfrac{2}{3}\times \dfrac{1}{2},x=\dfrac{-2}{2} \\
\end{align}$
Now, on simplifying the above equation we will get
$\Rightarrow x=\dfrac{1}{3},x=-1$
Hence on solving the given equation we get the values of x as $-1,\dfrac{1}{3}$.

Note: Alternatively we can solve the given equation as:
$\Rightarrow {{\left( x+\dfrac{1}{3} \right)}^{2}}-\dfrac{4}{9}=0$
We can rewrite the given equation as
$\Rightarrow {{\left( x+\dfrac{1}{3} \right)}^{2}}=\dfrac{4}{9}$
Now, taking the square root both sides we will get
$\Rightarrow \sqrt{{{\left( x+\dfrac{1}{3} \right)}^{2}}}=\sqrt{\dfrac{4}{9}}$
Now, simplifying the above obtained equation we will get
$\Rightarrow x+\dfrac{1}{3}=\pm \dfrac{2}{3}$
Now, simplifying the above obtained equation we will get
$\Rightarrow x+\dfrac{1}{3}=\dfrac{2}{3}$ and $x+\dfrac{1}{3}=-\dfrac{2}{3}$
$\Rightarrow x=\dfrac{2}{3}-\dfrac{1}{3}$ and $x=-\dfrac{2}{3}-\dfrac{1}{3}$
$\Rightarrow x=\dfrac{2-1}{3}$ and $ x=\dfrac{-2-1}{3}$
$\Rightarrow x=\dfrac{1}{3}$ and $ x=\dfrac{-3}{3}=-1$
Hence we get two values of x as $-1,\dfrac{1}{3}$ .