# SOLUTION: can you help me solve these quadratic equations by factoring.. m^2+8m+16=0, 6r^2-r-2=0, 9s^2+12s=-4, m^2-100=0 thank you

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: can you help me solve these quadratic equations by factoring.. m^2+8m+16=0, 6r^2-r-2=0, 9s^2+12s=-4, m^2-100=0 thank you      Log On

 Algebra: Linear Algebra (NOT Linear Equations) Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Linear Algebra Question 30142: can you help me solve these quadratic equations by factoring.. m^2+8m+16=0, 6r^2-r-2=0, 9s^2+12s=-4, m^2-100=0 thank youAnswer by sdmmadam@yahoo.com(530)   (Show Source): You can put this solution on YOUR website!can you help me solve these quadratic equations by factoring.. m^2+8m+16=0, 6r^2-r-2=0, 9s^2+12s=-4, m^2-100=0 thank you 1)m^2+8m+16=0 ----(1) This is a perfect square = 0 (m)^2 + 2X(m)X(4) +(4)^2 = 0 (which is like a^2 +2ab +b^2 and hence = (a+b)^2 and here a = m and b = 4) (m+4)^2 = 0 (m+4)(m+4) =0 (m+4) = 0 gives m= -4 (1) is an equation with the root (-4) occuring again. We say that (-4) is the root with multiplicity 2 Verification: m= -4 in (1) LHS = m^2+8m+16 = (-4)^2 + 8X(-4) +16 = 16-32+16 = (32-32) =0 = RHS 2)6r^2-r-2=0 ----(1) 6r^2-4r+3r-2=0 (the mid term (-r) is expressed as (-4r+3r) so that (-4r)X(3r) = -12r^2 = (6r^2)X(-2) ) (that is sum is (-1) and the product is (-12) and hence the quantities are (-4) and (3) ) (6r^2-4r)+(3r-2)=0 (by additive associativity) 2r(3r-2)+1(3r-2) =0 2rp +p =0 where p= (3r-2) p(2r+1) =0 (3r-2)(2r+1) = 0 (putting p back) (3r-2) =0 gives 3r = 2 implying r = (2/3) (2r+1) = 0 gives 2r = -1 implying r = (-1/2) Answer: r = 2/3 and r =(-1/2) Verification: r = 2/3 in (1) LHS = 6r^2-r-2 = 6X(4/9)-(2/3)-2 = (1/9)X(24-6-18) = (1/9)X(0) = 0 = RHS r = (-1/2) in (1) 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 Therefore our values are correct. 3)9s^2+12s=-4 9s^2+12s+ 4 =0 ----(1) This is a perfect square = 0 (3s)^2 +2X(3s)X(2) + (2)^2 =0 (which is like a^2 +2ab +b^2 and hence = (a+b)^2 and here a = 3s and b = 2) (3s+2)^2 = 0 (3s+2)=0 gives 3s = -2 which implies s = (-2/3) (1) is an equation with the root (-2/3) occuring again. We say that (-2/3) is the root with multiplicity 2 Verification: s= (-2/3) in (1) 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 4)m^2-100=0 (m)^2-(10)^2 = 0 (which is like a^2-b^2 = (a+b)(a-b) ) (m+10)(m-10)= 0 (m+10) = 0 gives m= -10 (m-10) = 0 gives m = 10 Verification: Both (10)^2 and (-10)^2 give 100