Complete the square x^2 + 12x + 9 - 6x + 5
Enter your answer in the form a(x + u)^2 + v, where a, u, and v are replaced by numbers.
+1622 The first step is to combine like terms in the original expression in the question to make the quadratic clearer:
\(x^2 + 6x + 14\)
Now we must open the parenthesis for (x + u)^2.
\((x + u)^2\) = \(x^2 + 2xu + u^2\)
\(u\) is a constant so \(u^2\) is also a constant, meaning that \(2xu\) is the only term that can equal \(6x\) in the expression in the question.
Therefore the constant \(u = 3\).
If \(u\) is 3, then \(u^2\) would be 9, making the new expression:
\(ax^2 + 6ax + 9a + v\)
(don't forget about the variable \(a\) that multiplies the outcome of opening the parenthesis for (x + u)^2)
If you take a look at the expression the problem gave to us, the coefficient of \(x^2\) was 1, so that means the constant \(a\) is also 1.
The new expression would look like this and we can get the last constant we need:
\(x^2 + 6x + 9 + v\)
Now we see there is a difference of 5 from the expression above and the expression in the problem, so \(v = 5\).
Therefore the answer for this questions is:
\(1(x + 3)^2 + 5\)
where \(a = 1\), \(u = 3\), and \(v = 5\)
