How do you solve #-4r^2+21r=r+13# by completing the square?

3 Answers
Oct 19, 2017

Answer:

#r_1=(5+2sqrt(3))/2# and #r_2=(5-2sqrt(3))/2#

Explanation:

#-4r^2+21r=r+13#

#4r^2-21r+r+13=0#

#4r^2-20r+13=0#

#4r^2-20r+25-12=0#

#(2r-5)^2-(2sqrt(3))^2=0#

#(2r-5+2sqrt(3))*(2r-5-2sqrt(3))=0#

Hence #r_1=(5+2sqrt(3))/2# and #r_2=(5-2sqrt(3))/2#

Oct 26, 2017

Answer:

#r = 4.232" "or r = 0.768#

Explanation:

#-4r^2+21r =r+13" "larr# make #r^2# term positive:

#0 = 4r^2 -21r +r+13" "larr#simplify

#4r^2 -20r +13=0" "larr# now #div4#

#r^2 -5r +13/4 =0" "larr# move the constant to the RHS

#r^2 -5r+color(blue)(???) = -13/4+color(blue)(???)" "larr# add #color(blue)((b/2)^2)# to both sides

#r^2 -5r +color(blue)((-5/2))^2 = -13/4 +color(blue)((-5/2)^2)#

#" "(r-5/2)^2 = -13/4+color(blue)(25/4) = 12/4 =3#

#" "(r-5/2)^2 =3" "larr# find square root of both sides

#r-5/2 = +-sqrt3#

#r = +sqrt3+2.5" "or r = -sqrt3+2.5#

#r = 4.232" "or r = 0.768#

Oct 26, 2017

Answer:

#r= 5/2-sqrt3 or r=5/2+sqrt3#

Explanation:

#-4r^2 +21r =r+13#
Let's start by subtracting #color(red)(r+13)# from both sides
#-4r^2 +21r - color(red)(r+13)=cancel(r+13)cancelcolor(red)(-r-13)#
#4r^2 + 21r -r - 13= 0#
#4r^2 +20r -13=0#
Now we can use the quadratic formula
We have:
#a=-4#
#b=20#
#c=-13#
Note: The quadratic formula is a formula that is very important and I recommend you to memorize it. At least understand how and when to use it.
#r = ((-b)+-sqrt(b^2-4ac))/(2a)#
Replace the numbers
#r = ((-20)+-sqrt((20)^2-4(-4)(-13)))/((2)(-4))#
#r=((-20)+-sqrt(192))/-8#
#r= 5/2 -sqrt3 or r=5/2 +sqrt3#