#### Question

# How would you solve 25x^2-60x=-31 by "completing the square"?

The answer should be (6(+/-)sqrt5)/5. Not sure how to get it though.

#### Answer

25x^2 - 60x = -31

x^2 - 60x/25 = -31/25

x^2 - 12x/5 = -31/25

x^2 - 12x/5 + [(-12/5) / 2]^2 = -31/25 + [(-12/5) / 2]^2

x^2 - 12x/5 + 144/100 = -31/25 + 144/100

x^2 - 12x/5 + 36/25 = -31/25 + 36/25

(x - 6/5)(x - 6/5) = 5/25

(x - 6/5)^2 = 5/25

x - 6/5 = ± √(5 / 25)

x - 6/5 = ± √5 / 5

x = 6/5 ± √5/5

x = (6 ± √5) / 5

you're teacher is mean

this type of quadratic is solved easier with the quadratic formula

25x^2 - 60x + 31 = 0

a = 25

b = -60

c = 31

[-b ± √(b^2 - 4ac)] / 2a

(60 ± √500) / 50

(60 ± 10√5) / 50

x = (6 ± √5) / 5

simple 25x^2-60x+31=( (5x)^2-2(5)(6)x+(6)^2-5=(5x-6)^2-5. That is (5x-6)^2=5

Then 5x-6= sqrt(5) or 5x-6=-sqrt(5). That is, x= (6+-sqrt(5))/5

25x2 -60x= -31

(5x)2 -2*6*(5x)= -31

To complete the square you need to add 62 =36 to both sides

(5x)2 -2*6*(5x) +62 = 36 -31

(5x -6)2 = 5

5x- 6 = +/-√5

5x = 6 +/-√5

x = ( 6 +/-√5)/5

25x^2-60x+31=0

25{x^2-(12/5)x+(36/25)}+31-36=0

this is the confusing step. if u write it out on paper, it should be more clearer

25(x-6/5)^2=5

(x-6/5)^2=1/5

*square root everything

x-6/5=(+or-)1/(root 5)

x=(6(+/-)(root5))/5