How do you solve #sqrt(4y+21)=y#?
2 Answers
Explanation:
squaring on both sides
rearranging the equation
Then we have to check if both these values of
1: When
which is true,
2:When
Case-1
Therefore, the value of
Case-2
-3 = -3
Hence 7 and -3 are roots for this equation
Explanation:
When
case -1
-3 = -3
=0
Hence -3 also satisfied the equation, hence -3 is also root for the above equation along with 7