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

Oct 15, 2015

$y$ = $7$

#### Explanation:

$\sqrt{4 y + 21}$ = $y$

squaring on both sides

${\left(\sqrt{4 y + 21}\right)}^{2}$ =${y}^{2}$
$4 y + 21 = {y}^{2}$

rearranging the equation

${y}^{2} - 4 y - 21 = 0$

${y}^{2} - 7 y + 3 y - 21 = 0$
$y \left(y - 7\right) + 3 \left(y - 7\right) = 0$
$\left(y - 7\right) \left(y + 3\right) = 0$
$y - 7 = 0 \mathmr{and} y + 3 = 0$
$y = 7 \mathmr{and} y = - 3$

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

1: When $y = 7$,

$\sqrt{4 \times 7 + 21}$ = $7$
$\mathmr{and} , \sqrt{28 + 21}$ = $7$
$\mathmr{and} , \sqrt{49}$ = $7$
$\mathmr{and} , 7 = 7$

which is true, $\therefore y = 7$

2:When $y = - 3$,
Case-1
$\sqrt{4 \times \left(- 3\right) + 21}$ = $- 3$
$\mathmr{and} , \sqrt{- 12 + 21}$ = $- 3$
$\mathmr{and} , \sqrt{9}$ = $- 3$
$\mathmr{and} , 3 \ne - 3$

$\therefore y \ne - 3$

Therefore, the value of $y$ is $7$
Case-2
$\sqrt{4 \times \left(- 3\right) + 21}$ = $- 3$
$\mathmr{and} , \sqrt{- 12 + 21}$ = $- 3$
$\mathmr{and} , \sqrt{9}$ = $- 3$
$\mathmr{and} , \sqrt{- {3}^{2}} =$-3#
-3 = -3

Hence 7 and -3 are roots for this equation

Oct 17, 2015

$y = 7 \mathmr{and} - 3$

#### Explanation:

$\sqrt{4 y + 21} = y$

When $y$= -3

$\sqrt{4 X \left(- 3\right) + 21} = - 3$
$\sqrt{- 12 + 21} = - 3$
$\sqrt{9} = - 3$

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