(y-2)(y+3) - (y-1)^2 = 0Expanding the left side, we get:y^2 + 3y - 2y - 6 - (y^2 - 2y + 1) = 0y^2 + y - 6 - y^2 + 2y - 1 = 03y - 7 = 03y = 7y = 7/3
(p+1)^2 - (p+2)^2 = 0Expanding the left side, we get:p^2 + 2p + 1 - (p^2 + 4p + 4) = 0p^2 + 2p + 1 - p^2 - 4p - 4 = 0-2p - 3 = 0-2p = 3p = -3/2
Therefore, the solutions to the given biquadratic equations are:
(y-2)(y+3) - (y-1)^2 = 0
Expanding the left side, we get:
y^2 + 3y - 2y - 6 - (y^2 - 2y + 1) = 0
y^2 + y - 6 - y^2 + 2y - 1 = 0
3y - 7 = 0
3y = 7
y = 7/3
(p+1)^2 - (p+2)^2 = 0
Expanding the left side, we get:
p^2 + 2p + 1 - (p^2 + 4p + 4) = 0
p^2 + 2p + 1 - p^2 - 4p - 4 = 0
-2p - 3 = 0
-2p = 3
p = -3/2
Therefore, the solutions to the given biquadratic equations are:
y = 7/3p = -3/2