By completing the square, find the zeros of y = x ^2 + 8x + 5?

Answer 1

Before we start with the main problem, let's work out how to get a perfect square with "x^2 + 8x"

Remember, a perfect square is of the form:
x^2 + bx + (b/2)^2 = (x - b/2) ^2

So, all you have to do is identify the "b" value, then divide it by 2, and then square it, to find the value that will "complete the square"

For x^2 + 8x, b = 8... So the value that will complete the square will be (8/2)^2 = 4^2 = 16.

Keep that in mind...

NOW, let's get to the main problem:
y = x ^2 + 8x + 5

If we want to complete the square, then we want to add 16 to the equation... but we can't just randomly add a number to an equation w/o messing it up, so we also have to subtract it at the same time, so that the overall change is 0

y = x^2 + 8x + 16 + 5 - 16
y = (x^2 + 8x + 16) + 5 - 16

Now, remember how to factor that perfect square in the paranthesis? (Look back at the top if you don't)

y = (x + 4)^2 + 5 - 16
y = (x + 4)^2 - 11

Okay... so now to solve for the zeros.

Set the equation to 0
0 = (x+4)^2 - 11
11 = (x+4)^2

When you square it, you get a positive root and a negative root. So, now you have:
(x+4) = + √11
(x+4) = - √11

So you also have 2 answers for x:
x = -4 + √11
x = -4 - √11

Answer 2

y = x^2 + 8x + 5

The zeros are found by making y = 0 and solving for x.

0 = x^2 + 8x + 5

Complete the square. Add and subtract 16.

0 = x^2 + 8x + 16 + 5 - 16

Factor the now-square trinomial.

0 = (x + 4)^2 + 5 - 16
0 = (x + 4)^2 - 11

Move the -11 to the left hand side.

11 = (x + 4)^2

Take the square root of both sides, keeping in mind that we have to include +/- when doing so

+/- sqrt(11) = x + 4

-4 +/- sqrt(11) = x

Our two zeros are

x = { -4 + sqrt(11) , -4 - sqrt(11) }

Answer 3

x^2 + 8x + (11+5) = 11

(x + 4)^2 = 11

Take the square root of both sides

x + 4 = 11^0.5

x = 11^0.5 - 4