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.

Guest Jan 23, 2022

#1**+2 **

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 \(v = 5\)

proyaop Jan 23, 2022