SOLUTION: Solve by completing the square: I attempted to do this problem, can you please analyze and inform me as to where I may have gone wrong. I think I missed something or did a step wro

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Question 63911This question is from textbook intermediate algebra
: Solve by completing the square: I attempted to do this problem, can you please analyze and inform me as to where I may have gone wrong. I think I missed something or did a step wrong.
x^2+10x+13=0
x^2+10x+13-13=0-13
x^2+10x+(5)^2=-13+(5)^2
(x+5)^2=-13+25
(x+5)^2=5
x+5=±5
x+5-5=-5±5
x=-5±5
x=-5-5
x=-10
or
x=-5+5
x=-0 This question is from textbook intermediate algebra

Found 3 solutions by venugopalramana, joyofmath, tanimachatterjee:
Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!
SEE MY COMMENTS BELOW
Solve by completing the square: I attempted to do this problem, can you please
analyze and inform me as to where I may have gone wrong. I think I missed something or did a step wrong.
x^2+10x+13=0.......OK
x^2+10x+13-13=0-13.........OK
x^2+10x+(5)^2=-13+(5)^2................GOOD
(x+5)^2=-13+25................GOOD
(x+5)^2=5....NO...ON RHS 25-13=12...NOT...5
x+5=±5.....NO... X+5 = + OR - SQRT(12)
x+5-5=-5±5..NO...X=-5+SQRT(12)=-1.5359...OR...-5-SQRT(12)=-8.4641
x=-5±5
x=-5-5
x=-10
or
x=-5+5
x=-0
Answer by joyofmath(189)   (Show Source): You can put this solution on YOUR website!

Add 12 to both sides: .
Rewrite as .
Take the square root of both sides: .
So, or . But, can be written as so the two values of x are .
Verify these values using the quadratic formula:
where a=1, b=10, and c=13.
Then .
Answer by tanimachatterjee(60)   (Show Source): You can put this solution on YOUR website!
x^2+10x+13=0
x^2+10x+13-13=0-13
x^2+10x+(5)^2=-13+(5)^2
(x+5)^2=-13+25
after that
u did some mistakes
(x+5)^2=12

now x+5 = (12)^(1/2)=2* (3)^(1/2) =3.46
x= 3.46 - 5 =-1.5

or x+5 = -(12)^(1/2)=-2 * (3)^(1/2)=-3.46
x=-3.46 -5=-8.46
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