How do you solve p^2 + 2p + 1 =0?

3 Answers
Konstantinos Michailidis
May 31, 2016

It is

p^2+2p+1=0=>p^2+p+p+1=0=>p(p+1)+(p+1)=0=> (p+1)(p+1)=0=>(p+1)^2=0

Hence (p+1)^2=0=>p+1=0=>p=-1

David B.
May 31, 2016

There are two ways to solve for p in this case 1) factoring and 2) using the quadratic formula. Both yield p=-1

Explanation:

There are two ways to solve for p in this case 1) factoring and 2) using the quadratic formula.

1) Factoring

We are looking for a way to re-write the left-hand side of this equation as the multiplication of two factors, i.e.

a(p+b)(p+c)=p^2+2p+1 = 0

We can see immediately that a=1. We are then left with:

p^2 + (b+c)p +bc = p^2+2p+1 = 0

The solution of this is letting b=c=1, therefore we have:

(p+1)(p+1) = 0

Therefore the solution is p=-1

2) Quadratic formula

To use this formula we need the quadratic in the form

0=ap^2+bp+c

Using this we see that a=1, b=2, and c=1. The formula is then

p=(-b+-sqrt(b^2-4ac))/(2a)

p=(-2-sqrt(4-4))/(2) = -1

Don't Memorise
May 31, 2016

The solution for the equation is:
color(blue)(p=-1

Explanation:

p^2 +2p +1 = 0

The equation is of the form color(blue)(ap^2+bp+c=0 where:

a=1, b=2, c=1

The Discriminant is given by:

Delta=b^2-4*a*c

= (2)^2-(4* 1 * 1)

= 4-4 = 0

The solutions are found using the formula
p=(-b+-sqrtDelta)/(2*a)

p = ((-2)+-sqrt(0))/(2*1) = (-2+-0)/2

p = ( -2 +0 ) /2 = -2/2 = -1

p = (-2 -0) /2 = -2/2 = -1

The solution for the equation is:
color(blue)(p=-1

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