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

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> 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      Log On


   

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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B10x%2B13=0
Add 12 to both sides: x%5E2%2B10x%2B25=12.
Rewrite as %28x%2B5%29%5E2=12.
Take the square root of both sides: x%2B5+=+%2B-+sqrt%2812%29.
So, x=-5%2Bsqrt%2812%29 or x=-5-sqrt%2812%29. But, sqrt%2812%29 can be written as sqrt%28%284%29%283%29%29+=+2sqrt%283%29 so the two values of x are -5+%2B-+2sqrt%283%29.
Verify these values using the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ where a=1, b=10, and c=13.
Then .

Answer by tanimachatterjee(60) About Me  (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