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

Question

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

anonymous · Answer

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

campbell_st · Answer

(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