How do you solve using the completing square method \[{x^2} + 8x + 16 = 36\] ?

Hint: We first make the coefficient of \[{x^2}\] as 1 by dividing the complete equation by the coefficient of \[{x^2}\] . 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 and by taking LCM we will get the desired result.

Complete step by step solution:
Given, \[{x^2} + 8x + 16 = 36\] .
We need the coefficient of \[{x^2}\] as 1, we have 1 so no need to change,
 \[{x^2} + 8x + 16 = 36\]
The next step is we need to shift the constant terms to the right hand side of the equation,
 \[ \Rightarrow {x^2} + 8x = 36 - 16\] .
 \[ \Rightarrow {x^2} + 8x = 20{\text{ }} - - - (1)\]
Now we can see that the coefficient of ‘x’ is \[8\] . We divide the coefficient of ‘x’ by 2 and we square it.
 \[{\left( {\dfrac{8}{2}} \right)^2} = {(4)^2} = 16\] .
Now we need to add ‘16’ on both sides of the equation (1).
 \[ \Rightarrow {x^2} + 8x + 16 = 20 + 16\]
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 \[b = 4\] .
 \[ \Rightarrow {(x + 4)^2} = 36\]
Taking square root on both sides we have,
 \[ \Rightarrow x + 4 = \pm \sqrt {36} \]
 \[ \Rightarrow x + 4 = \pm 6\]
That is we have two roots,
 \[ \Rightarrow x + 4 = 6\] and \[x + 4 = - 6\]
 \[ \Rightarrow x = 6 - 4\] and \[x = - 6 - 4\]
 \[ \Rightarrow x = 2\] and \[x = - 10\] , is the required solution.
So, the correct answer is “ \[ x = 2\] and \[x = - 10\] ”.

Note: 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 side of the equation t0 another side of the equation, always change sign from positive to negative and vice-versa.