# SOLUTION: (7) Question: Solve for y: -2y-2 = sqrt 2y^2-7 -7 (You may include extraneous solutions) (A) 2, 8 (B) -2, 8 (C) -2, -8 (D) 2, -8 Please be specific as to which letter ans

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: (7) Question: Solve for y: -2y-2 = sqrt 2y^2-7 -7 (You may include extraneous solutions) (A) 2, 8 (B) -2, 8 (C) -2, -8 (D) 2, -8 Please be specific as to which letter ans      Log On

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 Question 30454: (7) Question: Solve for y: -2y-2 = sqrt 2y^2-7 -7 (You may include extraneous solutions) (A) 2, 8 (B) -2, 8 (C) -2, -8 (D) 2, -8 Please be specific as to which letter answer is correct- thanks!Answer by sdmmadam@yahoo.com(530)   (Show Source): You can put this solution on YOUR website!If the given problem is -2y-2 =sqrt(2y^2-7)-7 ----(1) -2y-2+7 = sqrt(2y^2-7) -2y+5 = sqrt(2y^2-7) Squaring both the sides (5-2y)^2 = (2y^2-7) [by additive commutativity -2y+5 = 5-2y and using [sqrt(p)]^2 = p ] 25+4y^2-20y = 2y^2-7 (using (a-b)^2 = a^2+b^2-2ab and here a = 5 and b = 2y ) (4y^2-2y^2)-20y+25+7 = 0 (grouping like terms,changing sign while changing side) 2y^2-20y+32 = 0 dividing by 2 y^2-10y+16 =0 y^2-8y-2y+16 = 0 (since sum is -10 and product is 16, the two terms into which the middle term is split are -8y and -2y so that -10y = (-8y)+(-2y) and (-8y)X(-2y) = 16y^2 = (y^2)X(16) ) y(y-8)-2(y-8) = 0 yp-2p = 0 where p= (y-8) p(y-2) =0 (y-8)(y-2) = 0 (y-8) = 0 gives y = 8 and (y-2) = 0 gives y = 2 Answer: y = 2 and y = 8 which is your choice (A) 2,8