Question

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

Answer 1

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

& \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{aligned}$$ Now, we can solve the given equation by using a quadratic formula. Then we will get $\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Now, on comparing the obtained equation with the general equation $a{{x}^{2}}+bx+c=0$we will get the values as $a=1,b=\dfrac{2}{3},c=\dfrac{-1}{3}$ Now, substituting the values we will get $\Rightarrow x=\dfrac{-\left( \dfrac{2}{3} \right)\pm \sqrt{{{\left( \dfrac{2}{3} \right)}^{2}}-4\times 1\times \dfrac{-1}{3}}}{2\times 1}$ Now, simplifying the above obtained equation we will get $$\begin{aligned} & \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{aligned}$$ 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 $\begin{aligned} & \Rightarrow x=\dfrac{\dfrac{-2+4}{3}}{2},x=\dfrac{\dfrac{-2-4}{3}}{2} \\\ & \Rightarrow x=\dfrac{\dfrac{2}{3}}{2},x=\dfrac{\dfrac{-6}{3}}{2} \\\ & \Rightarrow x=\dfrac{2}{3}\times \dfrac{1}{2},x=\dfrac{-2}{2} \\\ \end{aligned}$ 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 $-1,\dfrac{1}{3}$ .