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

Equation has no true solutions.

Given: $y + 1 = \sqrt(-2y-3)$

squaring both sides

${(y + 1)}^2$  =-2y-3

${y}^2 + {1}^2$+ 2y = -2y - 3

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

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

y = -2

TO VERIFY:

The law cos(x+90⁰)=-sin(x) can be proved by putting y=90⁰ in the following

Formula,

cos(x+y)=cos(x).cos(y)-sin(x).sin(y)

substitute value of y in equation

$-2+1 =\sqrt(-2(-2)-3)$ gives:

-1 = 1 which is incorrect.

Therefore, There are no solutions to the equation.

Alternatively,

If (a1/a2) = (b1/b2) ≠ (c1/c2), then there will be no solution. This type of system of equations is called an inconsistent pair of linear equations. If we plot the graph, the lines will be parallel and system of equations have no solution.

To know more, visit:

brainly.com/question/14721094

#SPJ4