Question

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

Answer 1

`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`

Answer 2

`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`

Answer 3

`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`