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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.

Respuesta :

calculista

Answer:

There are no true solutions to the equation.

Step-by-step explanation:

The correct equation is

[tex]y+1=\sqrt{-2y-3}[/tex]

Solve for y

squared both sides

[tex](y+1)^2=(-2y-3)[/tex]

[tex](y^2+2y+1)=(-2y-3)[/tex]

[tex]y^2+2y+1+2y+3=0[/tex]

[tex]y^2+4y+4=0[/tex]

[tex](y+2)(y+2)=0[/tex]

[tex]y=-2[/tex]

Verify

substitute the value of y in the original expression

[tex]-2+1=\sqrt{-2(-2)-3}[/tex]

[tex]-1=1[/tex] ----> is not true

therefore

There are no true solutions to the equation.

THE ANSWER IS D(there are no true solutions to the equation)

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