document.write( "Question 29170: solve by completing the square?\r
\n" ); document.write( "\n" ); document.write( "x^2+4x-5=0 \n" ); document.write( "
Algebra.Com's Answer #15971 by sdmmadam@yahoo.com(530)\"\" \"About 
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solve by completing the square?\r
\n" ); document.write( "\n" ); document.write( "x^2+4x-5=0 ----(*)
\n" ); document.write( "x^2+4x =5
\n" ); document.write( "(x)^2+2X(x)X(2) =5
\n" ); document.write( "(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)
\n" ); document.write( "(x)^2+2X(x)X(2)+4 =5+4
\n" ); document.write( "(x+2)^2 =9
\n" ); document.write( "Taking the square root,
\n" ); document.write( "(x+2) = + or minus 3
\n" ); document.write( "(x+2) = 3 or (x+2) = -3
\n" ); document.write( "That is x= 3-2 or x = -3-2
\n" ); document.write( "x=1 and x=-5
\n" ); document.write( "Ans: x=1 and x=-5
\n" ); document.write( "Verificaiton:x=1 in (*)
\n" ); document.write( "LHS= x^2+4x-5 = (1)^2+4X(1)-5 = 1+4-5 = 0 =RHS
\n" ); document.write( "x=-5 in (*)
\n" ); document.write( "LHS= x^2+4x-5 = (-5)^2+4X(-5)-5 = 25+-20-5 = 25-25=0 =RHS
\n" ); document.write( "Therefore our values are correct.\r
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