How do you solve the quadratic equation 2x^2+10x+10=02x2+10x+10=0?

2 Answers
Apr 12, 2018

x=-5/2+-1/2sqrt5x=52±125

Explanation:

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

rArr2(x^2+5x+5)=02(x2+5x+5)=0

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

"solve "x^2+5x+5=0" using the "color(blue)"quadratic formula"solve x2+5x+5=0 using the quadratic formula

•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)xx=b±b24ac2a

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

rArrx=(-5+-sqrt(25-20))/2=(-5+-sqrt5)/2x=5±25202=5±52

rArrx=-5/2+-1/2sqrt5larrcolor(red)"exact solutions"x=52±125exact solutions

Apr 12, 2018

The answer is
x=(-5/2)+-sqrt(5/4)x=(52)±54

Explanation:

  1. The first step is to take a factor of 2:
    2x^2+10x+10=02x2+10x+10=0
    2(x^2+5x+5)=02(x2+5x+5)=0
  2. Then divide both sides by 2:
    (2(x^2+5x+5))/2=0/22(x2+5x+5)2=02
    x^2+5x+5=0x2+5x+5=0
  3. Now subtract both sides by 5:
    x^2+5x+5-5=0-5x2+5x+55=05
    x^2+5x=-5x2+5x=5
  4. Now you have to complete the square:
    x^2+5x+(5/2)^2=-5+(5/2)^2x2+5x+(52)2=5+(52)2
  5. Now factorize the LHS (Left Hand Side):
    (x+5/2)^2=-5+(5/2)^2(x+52)2=5+(52)2
  6. Simplify the RHS (Right Hand Side):
    (x+5/2)^2=-5/1+5^2/2^2(x+52)2=51+5222
    (x+5/2)^2=(-5/1*4/4)+25/4(x+52)2=(5144)+254
    (x+5/2)^2=-20/4+25/4(x+52)2=204+254
    (x+5/2)^2=5/4(x+52)2=54
  7. Now solve for xx.
    x+5/2=+-sqrt(5/4)x+52=±54
    x=-5/2+-sqrt(5/4)x=52±54

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