How do you complete the square for ${x^2} + 20x$?

Hint: We have to find the complete square of a given polynomial by creating a trinomial square using algebraic identity. First, equate this polynomial with zero and make it an equation. Next, create a trinomial square on the left side of the equation. For this we have to find a value that is equal to the square of half of $b$. Next, add the term to each side of the given equation. Next, simplify the right-hand side of the equation by simplifying the term. Next, factor the perfect trinomial using (i).

Formula used: ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$……(i)
Where, $a$ and $b$ are any two numbers.

Complete step-by-step solution:
First, equate this polynomial with zero and make it an equation.
$ \Rightarrow {x^2} + 20x = 0$
Now, we have to create a trinomial square on the left side of the equation. For this we have to find a value that is equal to the square of half of $b$.
${\left( {\dfrac{b}{2}} \right)^2} = {\left( {\dfrac{{20}}{2}} \right)^2}$
Now, we have to add the term to each side of the given equation.
$ \Rightarrow {x^2} + 20x + {\left( {10} \right)^2} = {\left( {10} \right)^2}$
Now, we have to simplify the right-hand side of the equation by simplifying the term. For this raise $10$ to the power of $2$.
$ \Rightarrow {x^2} + 20x + {\left( {10} \right)^2} = 100$
Now, we have to factor the perfect trinomial into ${\left( {x + 10} \right)^2}$ using (i).
${\left( {x + 10} \right)^2} = 100$
Subtract $100$ from both sides of the equation, we get
$ \Rightarrow {\left( {x + 10} \right)^2} - 100 = 0$

Hence, with the help of formula (i) we obtain the complete square for ${x^2} + 20x$, ${\left( {x + 10} \right)^2} - 100$.

Note: We can also find the complete square of given polynomial by quadratic formula:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$……(ii)
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
Step by step solution:
First, equate this polynomial with zero and make it an equation.
$ \Rightarrow {x^2} + 20x = 0$
First, we have to compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} + 20x = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = 20$ and $c = 0$
Now, we have to substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( {20} \right)^2} - 4\left( 1 \right)\left( 0 \right)$
After simplifying the result, we get
$ \Rightarrow D = 400$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$ \Rightarrow x = \dfrac{{ - 20 \pm 20}}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$ \Rightarrow x = - 10 \pm 10$
$ \Rightarrow x + 10 = \pm 10$
 Square both sides of the equation, we get
$ \Rightarrow {\left( {x + 10} \right)^2} = 100$
$ \Rightarrow {\left( {x + 10} \right)^2} - 100 = 0$
Hence, with the help of formula (ii) we obtain the complete square for ${x^2} + 20x$, ${\left( {x + 10} \right)^2} - 100$.