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

Answer 1

There are no true solutions to the equation

Explanation:

The equation is

y+1=√-2y-3

Find y

Square both sides

(y+1)^2=(√-2y-3)^2

y^2+2y+1= -2y-3

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

y^2+4y+4=0

Solve quadratic equation by factorisation

we have

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

y=-2

Check

y+1=√-2y-3

-2+1=√-2(-2)-3

-1=√4-3

-1=√1

-1=1

This is not true