Question

How do you solve `sqrt(4y+21)=y`?

Answer 1

`y` = `7`

Explanation:

`sqrt(4y+21)` = `y`

squaring on both sides

`(sqrt(4y+21))^2` =`y^2`
`4y +21 = y^2`

rearranging the equation

`y^2-4y -21 = 0`

`y^2-7y+3y -21 = 0`
`y(y-7)+3(y-7) = 0`
`(y-7)(y+3) = 0`
`y-7 =0 or y+3 = 0`
`y=7 and y= -3`

Then we have to check if both these values of `y` will satisfy the original equation or not

1: When `y=7`,

`sqrt(4xx7+21)` = `7`
`or,sqrt(28+21)` = `7`
`or,sqrt(49)` = `7`
`or,7=7`

which is true, `therefore y=7`

2:When `y=-3`,
Case-1
`sqrt(4xx(-3)+21)` = `-3`
`or,sqrt(-12+21)` = `-3`
`or,sqrt(9)` = `-3`
`or,3!=-3`

`therefore y!=-3`

Therefore, the value of `y` is `7`
Case-2
`sqrt(4xx(-3)+21)` = `-3`
`or,sqrt(-12+21)` = `-3`
`or,sqrt(9)` = `-3`
`or, sqrt (-3^2)=`-3#
-3 = -3

Hence 7 and -3 are roots for this equation

Answer 2

`y = 7 and -3`

Explanation:

`sqrt(4y+21) = y`

When `y`= -3

`sqrt(4X(-3)+21) = -3`
`sqrt(-12+21)=-3`
`sqrt(9)=-3`

case -1

`sqrt(-3^2)= -3`

-3 = -3

=0
Hence -3 also satisfied the equation, hence -3 is also root for the above equation along with 7