Question

How do you solve ${{\left( 3x-2 \right)}^{2}}+54=0$?

Answer 1

Hint: In this problem we need to solve the given equation i.e., we need to calculate the value of $x$ from which the given equation is satisfied. In the equation we have ${{\left( 3x-2 \right)}^{2}}$ which is in the form of ${{\left( a-b \right)}^{2}}$. So, we will use the algebraic formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and simplify the given equation. Now we will get a quadratic equation which is in the form of $a{{x}^{2}}+bx+c=0$. To solve the obtained quadratic equation, we will use the quadratic formula which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We will substitute the values of $a$, $b$, $c$ in the above equation and simplify them to get the solution of the given equation.

Complete step by step answer:
Given that, ${{\left( 3x-2 \right)}^{2}}+54=0$.
Simplifying the term ${{\left( 3x-2 \right)}^{2}}$ by using the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, then the above equation is modified as
$\begin{align}
  & \Rightarrow {{\left( 3x \right)}^{2}}-2\left( 3x \right)\left( 2 \right)+{{2}^{2}}+54=0 \\
 & \Rightarrow 9{{x}^{2}}-12x+58=0 \\
\end{align}$
Comparing the above equation with $a{{x}^{2}}+bx+c=0$, then we will get
$a=9$, $b=-12$, $c=58$.
We have the formula for the solution of the quadratic equation $a{{x}^{2}}+bx+c=0$ as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of $a$, $b$, $c$ in the above equation, then we will get
$\Rightarrow x=\dfrac{-\left( -12 \right)\pm \sqrt{{{\left( -12 \right)}^{2}}-4\left( 9 \right)\left( 58 \right)}}{2\left( 9 \right)}$
When we multiply a negative sign with a negative sign, then we will get a positive sign. Using the above rule, we can write the above equation as
$\begin{align}
  & \Rightarrow x=\dfrac{12\pm \sqrt{144-2088}}{18} \\
 & \Rightarrow x=\dfrac{12\pm \sqrt{-1944}}{18} \\
\end{align}$
In the above equation we have the term $\sqrt{-1944}$. We can clearly say that the roots of the given equation contain imaginary numbers. Now the prime factorization of $1944$ will be $1944={{2}^{3}}\times {{3}^{5}}$. Now the value of $\sqrt{1944}=\sqrt{{{2}^{2}}\times 2\times {{9}^{2}}\times 3}=18\sqrt{6}$. Substituting this value in the above equation, then we will get
$\Rightarrow x=\dfrac{12\pm 18\sqrt{6}i}{18}$
Taking $6$ as common from the numerator of the above equation, then we will get
$\begin{align}
  & \Rightarrow x=\dfrac{6\left( 2\pm 3\sqrt{6}i \right)}{18} \\
 & \Rightarrow x=\dfrac{2\pm 3\sqrt{6}i}{3} \\
\end{align}$

Hence the solution of the given equation ${{\left( 3x-2 \right)}^{2}}+54=0$ is $x=\dfrac{2\pm 3\sqrt{6}i}{3}$

Note: It is advisable to use the quadratic formula for the solution of the quadratic equation since we don’t know whether the solution is imaginary or real. Getting an imaginary solution will be difficult in any other method like completing squares, factorizing. But in quadratic formula we can easily solve if the solution is imaginary or real.