$\sqrt {y+k}=y+1$ For what value of the constant $k$ does the given equation have $y$ as the only solution?

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Determine end behavior of polynomial and rational functions

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\sqrt {y+k}=y+1

For what value of the constant

k

does the given equation have

y

as the only solution?

Substitute and Simplify: Let's start by substituting y = $-1$ into the equation to find the value of k. $\newline$ $\sqrt{y+k} = y + 1$ $\newline$ Substitute y = $-1$ : $\newline$ $\sqrt{-1+k} = -1 + 1$
Square Both Sides: Now, simplify the right side of the equation: $\newline$ $\sqrt{-1+k} = 0$ $\newline$ To find the value of k, we need to square both sides of the equation: $\newline$ $(\sqrt{-1+k})^2 = 0^2$
Solve for k: After squaring both sides, we get: $\newline$ $-1 + k = 0$ $\newline$ Now, solve for k: $\newline$ k = $1$
Verify Solution: We need to verify that y = $-1$ is the only solution to the original equation with k = $1$ . Substitute k = $1$ back into the original equation: $\newline$ $\sqrt{y+1} = y + 1$ $\newline$ Now, let's check if there are any other solutions besides y = $-1$ .
Check for Other Solutions: For the equation $\sqrt{y+1} = y + 1$ to hold true, the right side must be non-negative because the square root function only yields non-negative results. Therefore, y + $1$ must be greater than or equal to $0$ : $\newline$ y + $1$ ≥ $0$ $\newline$ y ≥ $-1$
Finalize Solution: Since $y$ must be greater than or equal to $-1$ and we have already found that $y = -1$ is a solution, we need to ensure that there are no other solutions for $y \geq -1$ . If we try to find another solution where $y$ is greater than $-1$ , the right side of the equation will be greater than $1$ , but the left side will be the square root of a number greater than $1$ , which cannot equal the number itself. Therefore, $y = -1$ is the only solution.