Question

How do you complete the square of \[3{x^2} + 18x + 5\]?

Answer 1

Hint: We first make the coefficient of \[{x^2}\] as 1 by taking the coefficient of that variable. Add and subtract the square of half value of coefficient of x inside the bracket having rest of common terms. Use the identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] to write the bracket. Bring out the constant value and add or subtract that value with another constant value. Write the final equation with one complete square and rest constant terms.

Complete step-by-step answer:
We are given the quadratic equation \[3{x^2} + 18x + 5\] … (1)
We take out common the value in coefficient of \[{x^2}\]
\[ \Rightarrow 3({x^2} + 6x) + 5\] … (2)
Now we write the value that is to be added in the bracket i.e. half of coefficient of x and squaring the number obtained.
i.e. \[{\left( {\dfrac{6}{2}} \right)^2} = {3^2}\]
Add and subtract this value inside the bracket o equation (2)
\[ \Rightarrow 3({x^2} + 6x + {3^2} - {3^2}) + 5\]
Now we can group the terms making up the identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
\[ \Rightarrow 3\left( {\left\{ {{x^2} + 6x + {3^2}} \right\} - {3^2}} \right) + 5\]
\[ \Rightarrow 3\left( {{{(x + 3)}^2} - {3^2}} \right) + 5\]
Multiply the term outside the larger bracket to terms inside the larger bracket
\[ \Rightarrow 3{(x + 3)^2} - 27 + 5\]
Add the constant terms in the equation
\[ \Rightarrow 3{(x + 3)^2} - 22\]

\[\therefore \]Value of \[3{x^2} + 18x + 5\] after completing the square is \[3{(x + 3)^2} - 22\].

Note:
Many students make mistakes by squaring the value of b and then divide by 2 instead they have to first divide by 2 and then square it. Also, keep in mind we don’t have to find the roots of the equation so we don’t use the other method of completing square i.e. shifting constant to other side and adding the same values to both sides in order to complete the square and then finding the root by taking square root of the variable i.e. x.