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possible solution(s) for y

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by sanju09 » Thu Apr 16, 2009 3:00 am
If y = √ (3 y + 4) then the product of all possible solution(s) for y is
A. -4
B. -2
C. 0
D. 4
E. 6


IMO D
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
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by deagez » Thu Apr 16, 2009 3:08 am
for some reason I am getting A (-4) for this one not D

square both sides you get Y^2 = 3Y + 4

make equation equal to zero Y^2 -3Y - 4 = 0

factor (y-4)(y+1)=0

so y is either 4 or -1

(4)(-1) = -4
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by sanju09 » Thu Apr 16, 2009 3:22 am
deagez wrote:for some reason I am getting A (-4) for this one not D

square both sides you get Y^2 = 3Y + 4

make equation equal to zero Y^2 -3Y - 4 = 0

factor (y-4)(y+1)=0

so y is either 4 or -1

(4)(-1) = -4
Well done deagez, but I think that although (-1)^2 = 1, the square root symbol is reserved for the POSITIVE root only. Therefore, -1 is NOT a valid solution of the original equation.
The mind is everything. What you think you become. -Lord Buddha



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by gmat740 » Thu Apr 16, 2009 3:51 am
square both sides you get Y^2 = 3Y + 4

make equation equal to zero Y^2 -3Y - 4 = 0

factor (y-4)(y+1)=0

so y is either 4 or -1

(4)(-1) = -4
Well I learnt something in my high school which can be helpful for Quadratic Equations


ax^2 + bx + c = 0

so sum of roots = -b/a

product of roots = c/a

here in this question, a=1, b = -3 and c= -4

so there is no need to solve the entire quadratic eqn

Although I know the answer is not -4 because of the sq-rt.
I just posted this method for others to revise.

Hope it might of help to you all

Karan
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