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#

Related questions
Impact of this question
7121 views around the world
You can reuse this answer
Creative Commons License