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

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
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