2. Solve the equation p2 + 4p = 1 by completing the square

Question

I just need help understanding how to do this practice problem so I can get ready for a test.

Answer 1

These were really confusing to me at first, too, but what we want to do is turn the left side into something we can turn into, for example, (x - 5)². Here's how we do it.

The 1 ("c") is already on the other side for us, soooo we take half of "b", square it, and add it to both sides.

Half of 4 is 2, and 2 squared is 4, sooo...

p² + 4p + 4 = 1 + 4

Now, the easiest way to explain it is to take that half of "b" and put it inside with the parentheses.

(p + 2)² = 5

If we multiply (p + 2)(p + 2), we end up with p² + 4p + 4.

Now, we take the square root of both sides.

p + 2 = √5

p = √5 - 2

If you need to calculate further, use your calculator.

Answer 2

p² + 4p = 1

p² + 4p + (4/2)² = 1 + (4/2)²

p² + 4p + 4 = 1 + 4

(p + 2)² = 5

p + 2 = ±√5

p = -2 ±√5

Answer 3

You need to convert p^2+4p to ( p + some number)^2

To do this use 1/2 the coefficient of p which is this case is 4/2 =2

So we now have (p+2)^2. But in doing so we have added 4 to the left side of the equation ( p+2)^2 = p^2 +4p+4

So to keep things balanced substract 4 so now have (p+2)^2 -4 =1

(p+2)^ = 5

p+2 =+- sq root 5

p = -2 +-sqroot5