Questions: Solve for (y), where (y) is a real number. [ sqrt-8 y+9=y ] (If there is more than one solution, separate them with commas.)

Transcript text: Solve for $y$, where $y$ is a real number. \[ \sqrt{-8 y+9}=y \] (If there is more than one solution, separate them with commas.)

Solution

Step 1: Set Up the Equation

We start with the equation given in the problem: \[ \sqrt{-8y + 9} = y \]

Step 2: Square Both Sides

To eliminate the square root, we square both sides of the equation: \[ -8y + 9 = y^2 \]

Step 3: Rearrange the Equation

Rearranging the equation gives us a standard quadratic form: \[ y^2 + 8y - 9 = 0 \]

Step 4: Solve the Quadratic Equation

We can factor the quadratic equation: \[ (y + 9)(y - 1) = 0 \] This gives us the potential solutions: \[ y = -9 \quad \text{or} \quad y = 1 \]

Step 5: Check for Extraneous Solutions

We need to verify each solution in the original equation to ensure they are valid:

For $ y = -9 $: \[ \sqrt{-8(-9) + 9} = \sqrt{72 + 9} = \sqrt{81} = 9 \quad \text{(not equal to -9)} \] Thus, $ y = -9 $ is extraneous.
For $ y = 1 $: \[ \sqrt{-8(1) + 9} = \sqrt{-8 + 9} = \sqrt{1} = 1 \quad \text{(equal to 1)} \] Thus, $ y = 1 $ is a valid solution.

Step 6: Conclusion

The only valid solution to the equation is: \[ y = 1 \]