Question

Sebastian solved the radical equation y + 1 = but did not check his solution. (y + 1)2 = y2 + 2y + 1 = –2y – 3 y2 + 4y + 4 = 0 (y + 2)(y + 2) = 0 y = –2 Which is the true solution to the radical equation y + 1 = ? y = –2 y = 1 y = 2 There are no true solutions to the equation.

Answer 1

Answer:

There are no true solutions to the equation.

Step-by-step explanation:

The correct equation is

$y+1=\sqrt{-2y-3}$

Solve for y

squared both sides

$(y+1)^2=(-2y-3)$

$(y^2+2y+1)=(-2y-3)$

$y^2+2y+1+2y+3=0$

$y^2+4y+4=0$

$(y+2)(y+2)=0$

$y=-2$

Verify

substitute the value of y in the original expression

$-2+1=\sqrt{-2(-2)-3}$

$-1=1$ ----> is not true

therefore

There are no true solutions to the equation.