Question 29170: solve by completing the square?
x^2+4x-5=0
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! solve by completing the square?
x^2+4x-5=0 ----(*)
x^2+4x =5
(x)^2+2X(x)X(2) =5
(LHS resembles (a)^2+2ab where a=x and b=2. And we require (b)^2 to complete the square. So we add (b)^2 = 2^2 = 4 to both the sides)
(x)^2+2X(x)X(2)+4 =5+4
(x+2)^2 =9
Taking the square root,
(x+2) = + or minus 3
(x+2) = 3 or (x+2) = -3
That is x= 3-2 or x = -3-2
x=1 and x=-5
Ans: x=1 and x=-5
Verificaiton:x=1 in (*)
LHS= x^2+4x-5 = (1)^2+4X(1)-5 = 1+4-5 = 0 =RHS
x=-5 in (*)
LHS= x^2+4x-5 = (-5)^2+4X(-5)-5 = 25+-20-5 = 25-25=0 =RHS
Therefore our values are correct.
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