Question

How to solve the equation \[{\left( {4x - 5} \right)^2} = 64\].

Answer 1

Hint: Both the side square root can be taken to solve such questions. Since this is a quadratic equation so it will have two solutions. Each of the solutions when used in place of x then it must give the result that is here it must give 64. Hence substitution of values can be done in place of x that is LHS to verify the RHS.

Complete step by step answer:
First square root of both sides are taken
\[
\Rightarrow \sqrt {{{\left( {4x - 5} \right)}^2}} = \pm 64 \\
   \Rightarrow 4x - 5 = \pm 8 \\
 \]
Now this equation can be solved to get the roots or values for x that can be substituted.
Adding 5 both sides we get,
\[
\Rightarrow 4x - 5 + 5 = 8 + 5 \\
   \Rightarrow 4x = 13 \\
\]
Now, dividing both sides by 4 in the above equation we have,
\[
\Rightarrow \dfrac{{4x}}{4} = \dfrac{{13}}{4} \\
   \Rightarrow x = \dfrac{{13}}{4} \\
 \]
Now solving \[4x - 5 = - 8\]
\[
   \Rightarrow 4x - 5 + 5 = - 8 + 5 \\
   \Rightarrow 4x = - 3 \\
   \Rightarrow x = - \dfrac{3}{4} \\
 \]
As a check we can substitute these values into the left side and if equal to the right side then they are the solutions. Here both the values of x satisfy the equation.
Hence the values of x are \[ - \dfrac{3}{4}\],\[\dfrac{{13}}{4}\].

Note: In such questions a product of several terms equals zero. When a product of two or more terms equals zero then at least one of the terms must be zero. Each term equal to zero must be solved separately. In other words, the number of terms which are present, that many equations must be solved. Any solution of term equal to 0 solves product equal to 0 as well. Whatever solutions or roots are obtained after solving the equations can be verified by substituting them in the equations.