Question

How do you solve \[3{{\left( x-1 \right)}^{2}}=27\]?

Answer 1

Hint: To solve this question, we need to know an algebraic identity. We should know the expansion of algebraic expression \[{{a}^{2}}-{{b}^{2}}\]. The expansion of this expression is \[\left( a+b \right)\left( a-b \right)\]. For any general expression of this form, we can expand it by substituting the values of a and b for that expression. To express in the form \[{{a}^{2}}-{{b}^{2}}\], we need to take all terms to one side of the equation.

Complete step by step solution:
We are given the equation \[3{{\left( x-1 \right)}^{2}}=27\]. Dividing by 3 to both sides of the above equation, we get \[\dfrac{3{{\left( x-1 \right)}^{2}}}{3}=\dfrac{27}{3}\].
Cancelling out the common factors from both sides, we get
\[\Rightarrow {{\left( x-1 \right)}^{2}}=9\]
Subtracting 9 from both sides of the above equation, we get
\[\Rightarrow {{\left( x-1 \right)}^{2}}-9=0\]
We know that 9 is square of 3, using this we can express the right side as
\[\Rightarrow {{\left( x-1 \right)}^{2}}-{{3}^{2}}=0\]
The left side of the equation is of the form \[{{a}^{2}}-{{b}^{2}}\], we know that its expansion is \[\left( a+b \right)\left( a-b \right)\]. Hence, the left side of the equation is expanded as
\[\begin{align}
  & \Rightarrow \left( x-1+3 \right)\left( x-1-3 \right)=0 \\
 & \Rightarrow \left( x+2 \right)\left( x-4 \right)=0 \\
\end{align}\]
The above expression is the factored form of the equation. Hence, we get the roots of the equation as \[x=-2\] or \[x=4\].
Hence, the solutions of the given equation are \[x=-2\] or \[x=4\].

Note: We can check answers by substituting them in the equation.
Substituting \[x=-2\], in the equation, we get LHS as \[3{{\left( -2-1 \right)}^{2}}=3\times 9=27\], and RHS is 27. Thus, this solution is correct. Substituting \[x=4\] in the equation we get LHS as \[3{{\left( 4-1 \right)}^{2}}=3\times 9=27\] RHS is 27. Thus, this solution is also correct.