How do you solve #x^2+5x+2=0# by completing the square?

Question

20377 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-17T20:18:37

#x=-5/2+sqrt17/2# and #x=-5/2-sqrt17/2#

Half the middle term, add it to #x# and square it. Keep the constant on the end

#x^2+5x -> (x+5/2)^2+2#

Take away the term previously halved, but squared.

#(x+5/2)^2-25/4+8/4#

Turn the extra constant to have the same denominator to make simplifying easier

Simplify:

#(x+5/2)^2-17/4=0#

Solving:

Plus the constant, making it cancel out:

#(x+5/2)^2cancel(-17/4)=17/4#

Get rid of the square bracket by square rooting:

#x+5/2=pmsqrt(17/4)# #larr# Notice the #pm# sign, we add this when square rooting.

Minus the constant:

#x+cancel(5/2)=-5/2pmsqrt(17/4)#

Simplify (If needed):

#sqrt4 -> 2#

#x=-5/2pmsqrt17/2#

Therefore our 2 answers are:

#x=-5/2+sqrt17/2# and #x=-5/2-sqrt17/2#