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

2 Answers
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"

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.