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# How do you complete the square for ${{x}^{2}}+18x$?

Last updated date: 19th Sep 2024
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Hint: In this question we will add and subtract a term in the expression to complete the square for the expression. Completing the square implies that we have to convert the expression in the formula of ${{a}^{2}}+2ab+{{b}^{2}}$ so that it can be simplified and written as ${{(a+b)}^{2}}$minus the term we added to complete the square such that the value does not change.

We have the given expression as:
${{x}^{2}}+18x$
Now since the second term is positive in the expression, we will convert the expression into the formula of ${{(a+b)}^{2}}$.
We know that ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$but from the given expression we only have the first two terms of the formula therefore, we will deduce the third term.
Now from the above formula we can conclude that there is coefficient to the term ${{x}^{2}}$therefore, we can conclude that $a=x$.
Now the middle term in the expansion formula is $2ab$, since the second term is $18x$, we can simplify the term and write it as $2(9)x$. Now this is in the format of $2ab$ and since we already know that $x=a$, this implies that $b=9$.
Now the third term in the expansion is ${{b}^{2}}$. Since the value of $b=9$, on squaring the term, we get ${{b}^{2}}=81$.
Therefore, the expression can be written as:
$\Rightarrow {{x}^{2}}+18x+81$
And since we are adding the value $81$, we will subtract $81$ from the expression too. On subtracting, we get:
$\Rightarrow {{x}^{2}}+18x+81-81$
Now on completing the square, we get:
$\Rightarrow {{\left( x+9 \right)}^{2}}-81$, which is the required solution.

Note: It is to be remembered when a value is added and subtracted at the same time in an expression, its value does not change. Another formula for expansion should also be remembered while doing these types of sums which is ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$. The general formula for writing the third term should be remembered as $\dfrac{b}{2a}$.