Description
we can rearrange the equation as x^2+2*x*1+1^2=0
now, as we may know that (a+b)^2=a^2+2ab+b^2
so the above equation becomes (x+1)^2=0
now taking square root on both sides
x+1=0
subtracting 1 from both sides
x+1-1=0-1
or, x=-1
so the solution is x=-1
Explanation & Answer
we can check the number of solutions of equations in form ax^2 + b x+ c
by checking the discriminant= b^2 - 4 ac
here x^2 + 2 x +1 = 0
discriminant= b^2 - 4 ac
=2^2 - 4*1*1
=0
if discriminant = 0 it will have only one real solution...................
x^2 + 2 x +1 = 0
x^2 + x+x + 1=0
x(x+1) + 1(x+1) = 0
(x+1) (x+1) = 0
x + 1 = 0
x= -1 ................................................................ answer
hope you understood...please message if you have any doubts...thank you
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