Question

When the expression $-2x^2-20x-53$ is written in the form $a(x+d)^2+e$, where $a$, $d$, and $e$ are constants, then what is the sum $a+d+e$?

Answer 1

We start with $2 x^{2} -20x-53$ and wish to write it as $a(x+d) ^{2} +e$

First, pull 2 out from the first two terms: $2( x^{2} -10x)-53$

Let’s look at what is in parenthesis. In the final form this needs to be a perfect square. Right now we have $x^{2} -10x$ and we can obtain -10x by adding -5x and -5x. That is, we can build the following perfect square: $x^{2} -10x+25=(x-5) ^{2}$

The “problem” with what we just did is that we added to what was given. Let’s put the expression together. We have $2( x-5) ^{2}-53$ and when we multiply that out it does not give us what we started with. It gives us $2 x^{2} -20x+50-53=2 x^{2} -20x-3$

So you see our expression is not right. It should have a -53 but instead has a -3. So to correct it we need to subtract another 50.

We do this as follows: $2(x-5) ^{2}-53-50$ which gives us the final expression we seek:

$2(x-5) ^{2}-103$

If you multiply this out you will get the exact expression we were given. This means that:
a = 2
d = -5
e =  -103

We are asked for the sum of a, d and e which is 2 + (-5) + (-103) = -106