Questions: Solve for (y), where (y) is a real number. [ sqrt-y+30=y ] (If there is more than one solution, separate them with commas.)

Solution Steps

To solve the equation \(\sqrt{-y + 30} = y\), we need to isolate \(y\). We can start by squaring both sides to eliminate the square root, then solve the resulting quadratic equation.

We start with the given equation: \[ \sqrt{-y + 30} = y \]

To eliminate the square root, we square both sides of the equation: \[ (\sqrt{-y + 30})^2 = y^2 \] This simplifies to: \[ -y + 30 = y^2 \]

Rearrange the equation to form a standard quadratic equation: \[ y^2 + y - 30 = 0 \]

We solve the quadratic equation \(y^2 + y - 30 = 0\) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = 1\), and \(c = -30\). Plugging in these values, we get: \[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \] \[ y = \frac{-1 \pm \sqrt{1 + 120}}{2} \] \[ y = \frac{-1 \pm \sqrt{121}}{2} \] \[ y = \frac{-1 \pm 11}{2} \]

Solving for \(y\), we get two potential solutions: \[ y = \frac{-1 + 11}{2} = 5 \] \[ y = \frac{-1 - 11}{2} = -6 \]

We need to verify which of these solutions satisfy the original equation \(\sqrt{-y + 30} = y\):

• For \(y = 5\): \[ \sqrt{-5 + 30} = \sqrt{25} = 5 \] This is true.
• For \(y = -6\): \[ \sqrt{-(-6) + 30} = \sqrt{36} = 6 \] This does not equal \(-6\), so it is not a valid solution.

Final Answer