How do you solve the quadratic equation #2x^2+10x+10=0#?

2 Answers
Jim G.
Apr 12, 2018

#x=-5/2+-1/2sqrt5#

Explanation:

#"take out a "color(blue)"common factor "2#

#rArr2(x^2+5x+5)=0#

#"there are no whole number factors of + 5 which sum to + 5"#

#"solve "x^2+5x+5=0" using the "color(blue)"quadratic formula"#

#•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)#

#"with "a=1,b=5" and "c=5#

#rArrx=(-5+-sqrt(25-20))/2=(-5+-sqrt5)/2#

#rArrx=-5/2+-1/2sqrt5larrcolor(red)"exact solutions"#

Aneesh M.
Apr 12, 2018

The answer is
#x=(-5/2)+-sqrt(5/4)#

Explanation:

  1. The first step is to take a factor of 2:
    #2x^2+10x+10=0#
    #2(x^2+5x+5)=0#
  2. Then divide both sides by 2:
    #(2(x^2+5x+5))/2=0/2#
    #x^2+5x+5=0#
  3. Now subtract both sides by 5:
    #x^2+5x+5-5=0-5#
    #x^2+5x=-5#
  4. Now you have to complete the square:
    #x^2+5x+(5/2)^2=-5+(5/2)^2#
  5. Now factorize the LHS (Left Hand Side):
    #(x+5/2)^2=-5+(5/2)^2#
  6. Simplify the RHS (Right Hand Side):
    #(x+5/2)^2=-5/1+5^2/2^2#
    #(x+5/2)^2=(-5/1*4/4)+25/4#
    #(x+5/2)^2=-20/4+25/4#
    #(x+5/2)^2=5/4#
  7. Now solve for #x#.
    #x+5/2=+-sqrt(5/4)#
    #x=-5/2+-sqrt(5/4)#

There may be other ways to do it, but this is the way that I solved this.

Related questions
Impact of this question
2743 views around the world
You can reuse this answer
Creative Commons License