Ask Question
14 November, 14:25

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

+4
Answers (1)
  1. 14 November, 14:35
    0
    There are no true solutions to the equation

    Step-by-step 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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers