How do you solve the quadratic equation 2x^2+10x+10=02x2+10x+10=0?
2 Answers
Apr 12, 2018
Explanation:
"take out a "color(blue)"common factor "2take out a common factor 2
rArr2(x^2+5x+5)=0⇒2(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±√b2−4ac2a
"with "a=1,b=5" and "c=5with a=1,b=5 and c=5
rArrx=(-5+-sqrt(25-20))/2=(-5+-sqrt5)/2⇒x=−5±√25−202=−5±√52
rArrx=-5/2+-1/2sqrt5larrcolor(red)"exact solutions"⇒x=−52±12√5←exact solutions
Apr 12, 2018
The answer is
Explanation:
- 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 - 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 - Now subtract both sides by 5:
x^2+5x+5-5=0-5x2+5x+5−5=0−5
x^2+5x=-5x2+5x=−5 - Now you have to complete the square:
x^2+5x+(5/2)^2=-5+(5/2)^2x2+5x+(52)2=−5+(52)2 - Now factorize the LHS (Left Hand Side):
(x+5/2)^2=-5+(5/2)^2(x+52)2=−5+(52)2 - 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=(−51⋅44)+254
(x+5/2)^2=-20/4+25/4(x+52)2=−204+254
(x+5/2)^2=5/4(x+52)2=54 - 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.