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anonymous
 4 years ago
Solving Quadratic Equations by Completing the Square.
1) (x1)(x+5)=7
2) x(x6)27=0
3) x squared= 5(2x5)
4) x(2x+4)=0
Please solve for the x's and show a little work if possible! thx :D
anonymous
 4 years ago
Solving Quadratic Equations by Completing the Square. 1) (x1)(x+5)=7 2) x(x6)27=0 3) x squared= 5(2x5) 4) x(2x+4)=0 Please solve for the x's and show a little work if possible! thx :D

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(x1)(x+5)=7 x^2+4x12= 0 (x+6)(x2)=0 x= 6 or x= 2

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.1(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 (x3)^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
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