document.write( "Question 29903: Please would anybody help me solve this problem I would appreciate it very much
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\n" ); document.write( "\n" ); document.write( "-x^2+3x-2=0 \n" ); document.write( "
Algebra.Com's Answer #16684 by sdmmadam@yahoo.com(530)\"\" \"About 
You can put this solution on YOUR website!
-x^2+3x-2=0
\n" ); document.write( "Multiplying by (-1)
\n" ); document.write( "x^2-3x+2 = 0 ----(1)
\n" ); document.write( "x^2-2x-x+2 =0 [splitting the middle term and expressing it as the sum of two terms so that their product is the product of the square term and the constant term. Here (-3x) = (-2x)+(-x) and (-2x)X(-x) = 2x^2 = (x^2)X(2)]
\n" ); document.write( "(x^2-2x)-x+2 =0 (by additive associativity)
\n" ); document.write( "x(x-2)-1(x-2)=0 (why (-1) pulled out? so that to get(x-2)! )
\n" ); document.write( "xp-p= 0 where p= (x-2)
\n" ); document.write( "p(x-1) = 0 (taking p out)
\n" ); document.write( "(x-2)(x-1) = 0 (putting t back)
\n" ); document.write( "(x-2) = 0 gives x = 2
\n" ); document.write( "(x-1) = 0 gives x = 1
\n" ); document.write( "Answer: x = 1 and x = 2
\n" ); document.write( "Verification: x = 1 in (1)
\n" ); document.write( "LHS = 1-3+2 = 0 = RHS
\n" ); document.write( "x = 2 in (1)
\n" ); document.write( "LHS = 4-6+2 = 0=RHS\r
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