## anonymous 4 years ago Solving Quadratic Equations by Completing the Square. 1) (x-1)(x+5)=7 2) x(x-6)-27=0 3) x squared= 5(2x-5) 4) x(2x+4)=0 Please solve for the x's and show a little work if possible! thx :D

1. anonymous

(x-1)(x+5)=7 x^2+4x-12= 0 (x+6)(x-2)=0 x= -6 or x= 2

2. campbell_st

(1) expand... x^2 +4x -5 = 7 the constant on the LHS needs to be 4... so add 9 to both sides... x^2 + 4x +4 = 16 gives (x+2)^2 = 16 $(x+2) = \sqrt{16}$ $x = -2 \pm 4$ therefore x = 2, -6 (2) same process x^2 -6x = 27 add (-6/2)^2 to both sides.. x^2 - 6x +9 = 36 (x-3)^2 = 36 I'll let you solve it (3) x^2 = 10x - 25 or x^2 -10x + 25 = 0 is (x - 5)^2 = 0 1 solution (4) 2x^2 +4x = 0 or x^2 +2x = 0 (2/2)^1 is added to both sides... x^2 +2x + 1 = 1 (x +1)^2 = 1 /// you can finish it