possible solution(s) for y

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

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.

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