document.write( "Question 30142: can you help me solve these quadratic equations by factoring.. m^2+8m+16=0,
\n" ); document.write( "6r^2-r-2=0, 9s^2+12s=-4, m^2-100=0 thank you
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Algebra.Com's Answer #17184 by sdmmadam@yahoo.com(530)\"\" \"About 
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can you help me solve these quadratic equations by factoring.. m^2+8m+16=0,
\n" ); document.write( "6r^2-r-2=0, 9s^2+12s=-4, m^2-100=0 thank you
\n" ); document.write( "1)m^2+8m+16=0 ----(1)
\n" ); document.write( "This is a perfect square = 0
\n" ); document.write( "(m)^2 + 2X(m)X(4) +(4)^2 = 0
\n" ); document.write( "(which is like a^2 +2ab +b^2 and hence = (a+b)^2 and here a = m and b = 4)
\n" ); document.write( "(m+4)^2 = 0
\n" ); document.write( "(m+4)(m+4) =0
\n" ); document.write( "(m+4) = 0 gives m= -4
\n" ); document.write( "(1) is an equation with the root (-4) occuring again. We say that (-4) is the root with multiplicity 2
\n" ); document.write( "Verification: m= -4 in (1)
\n" ); document.write( "LHS = m^2+8m+16 = (-4)^2 + 8X(-4) +16 = 16-32+16 = (32-32) =0 = RHS\r
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\n" ); document.write( "\n" ); document.write( "2)6r^2-r-2=0 ----(1)
\n" ); document.write( "6r^2-4r+3r-2=0
\n" ); document.write( "(the mid term (-r) is expressed as (-4r+3r) so that (-4r)X(3r) = -12r^2 = (6r^2)X(-2) )
\n" ); document.write( "(that is sum is (-1) and the product is (-12) and hence the quantities are (-4) and (3) )
\n" ); document.write( "(6r^2-4r)+(3r-2)=0 (by additive associativity)
\n" ); document.write( "2r(3r-2)+1(3r-2) =0
\n" ); document.write( "2rp +p =0 where p= (3r-2)
\n" ); document.write( "p(2r+1) =0
\n" ); document.write( "(3r-2)(2r+1) = 0 (putting p back)
\n" ); document.write( "(3r-2) =0 gives 3r = 2 implying r = (2/3)
\n" ); document.write( "(2r+1) = 0 gives 2r = -1 implying r = (-1/2)
\n" ); document.write( "Answer: r = 2/3 and r =(-1/2)
\n" ); document.write( "Verification: r = 2/3 in (1)
\n" ); document.write( "LHS = 6r^2-r-2 = 6X(4/9)-(2/3)-2 = (1/9)X(24-6-18) = (1/9)X(0) = 0 = RHS
\n" ); document.write( " r = (-1/2) in (1)
\n" ); document.write( "LHS = 6r^2-r-2 = 6X(1/4)-(-1/2)-2 = 6/4+1/2-2= (3/2+1/2)-2 = 2-2 =0= RHS
\n" ); document.write( "Therefore our values are correct.\r
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\n" ); document.write( "\n" ); document.write( "3)9s^2+12s=-4
\n" ); document.write( "9s^2+12s+ 4 =0 ----(1)
\n" ); document.write( "This is a perfect square = 0
\n" ); document.write( "(3s)^2 +2X(3s)X(2) + (2)^2 =0
\n" ); document.write( "(which is like a^2 +2ab +b^2 and hence = (a+b)^2 and here a = 3s and b = 2)
\n" ); document.write( "(3s+2)^2 = 0
\n" ); document.write( "(3s+2)=0 gives 3s = -2 which implies s = (-2/3)
\n" ); document.write( "(1) is an equation with the root (-2/3) occuring again. We say that (-2/3) is the root with multiplicity 2
\n" ); document.write( "Verification: s= (-2/3) in (1)
\n" ); document.write( "LHS =9s^2+12s+ 4 = 9(-2/3)^2+12X(-2/3)+4=9X(4/9)-8+4=4-8+4=8-8=0=RHS\r
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\n" ); document.write( "\n" ); document.write( "4)m^2-100=0
\n" ); document.write( "(m)^2-(10)^2 = 0 (which is like a^2-b^2 = (a+b)(a-b) )
\n" ); document.write( "(m+10)(m-10)= 0
\n" ); document.write( "(m+10) = 0 gives m= -10
\n" ); document.write( "(m-10) = 0 gives m = 10
\n" ); document.write( "Verification: Both (10)^2 and (-10)^2 give 100
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