How do you solve the quadratic equation #2x^2+10x+10=0#?
2 Answers
Apr 12, 2018
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
Explanation:
- The first step is to take a factor of 2:
#2x^2+10x+10=0#
#2(x^2+5x+5)=0# - Then divide both sides by 2:
#(2(x^2+5x+5))/2=0/2#
#x^2+5x+5=0# - Now subtract both sides by 5:
#x^2+5x+5-5=0-5#
#x^2+5x=-5# - Now you have to complete the square:
#x^2+5x+(5/2)^2=-5+(5/2)^2# - Now factorize the LHS (Left Hand Side):
#(x+5/2)^2=-5+(5/2)^2# - 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# - 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.