Question

How would one complete the square \[{x^2} + 6x + \_\] ?

Answer 1

Hint: We first make the coefficient of \[{x^2}\] as 1 by dividing the complete equation by the coefficient of \[{x^2}\]. If there is any constant present in the given expression then shift the constant value to the right hand side of the equation. Add the square of half value of coefficient of ‘x’ on both sides of the equation. Afterwards we can simplify this using some simple algebraic identity \[{(a - b)^2} = {a^2} - 2ab + {b^2}\] to get the desired result.

Complete step by step solution:
Given, \[{x^2} + 6x + \_\].
We can see that the coefficient of \[{x^2}\]as 1. So no need to divide the equation by the coefficient of \[{x^2}\].
If we compare this with \[a{x^2} + bx + c\] we don’t have the value of ‘c’.
Now we can see that the coefficient of ‘x’ is \[6\]. We divide the coefficient of ‘x’ by 2 and we square it.
\[{\left( {\dfrac{6}{2}} \right)^2} = {(3)^2} = 9\].
Add 9 to the given expression we have,
\[ = {x^2} + 6x + 9\]
We know the algebraic identity \[{(a + b)^2} = {a^2} + 2ab + {b^2}\]. Comparing this with the left hand side of an equation we have \[a = x\] and \[b = 3\].
\[ \Rightarrow {x^2} + 6x + 9 = {(x + 3)^2}\]
This is the required answer.

Note: Completing the square is one of the methods for finding the roots of the quadratic polynomial. We can find the roots of the obtained result. Since we have expressed it in terms of squares. Then we have,
\[{(x + 3)^2} = 0\]
\[(x + 3)(x + 3) = 0\]
Using zero product principle we have
\[x + 3 = 0\] and \[x + 3 = 0\]
\[x = - 3\] and \[x = - 3\].
Hence the roots of \[{x^2} + 6x + 9\] are \[ - 3\] and $-3$.
Since we have a polynomial of degree two and hence it is called quadratic polynomial. If we have a polynomial of degree ‘n’ then we have ‘n’ roots. In the given problem we have a degree that is equal to 2. Hence the number of roots are 2. Also keep in mind when shifting values from one \[ - 3\] side of the equation t0 another side of the equation, always change sign from positive to negative and vice-versa.