Time remaining:
find all values of x that satifies x^2+2x+1=0

Mathematics
Tutor: None Selected Time limit: 0 Hours

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

Apr 23rd, 2015

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

Apr 23rd, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Apr 23rd, 2015
...
Apr 23rd, 2015
Dec 4th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer