Solve Square Root 56Bd09

1. **State the problem:** Solve for $y$ in the equation $$\sqrt{25 - 3y} = y + 1$$. 2. **Recall the domain restriction:** Since the expression under the square root must be non-negative, we have $$25 - 3y \geq 0 \implies y \leq \frac{25}{3}.$$ Also, the right side $y + 1$ must be non-negative because the square root is always non-negative, so $$y + 1 \geq 0 \implies y \geq -1.$$ Thus, the domain for $y$ is $$-1 \leq y \leq \frac{25}{3}.$$ 3. **Square both sides to eliminate the square root:** $$\left(\sqrt{25 - 3y}\right)^2 = (y + 1)^2$$ which simplifies to $$25 - 3y = (y + 1)^2.$$ 4. **Expand the right side:** $$(y + 1)^2 = y^2 + 2y + 1.$$ So the equation becomes $$25 - 3y = y^2 + 2y + 1.$$ 5. **Bring all terms to one side to form a quadratic equation:** $$0 = y^2 + 2y + 1 + 3y - 25$$ which simplifies to $$0 = y^2 + 5y - 24.$$ 6. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=5$, and $c=-24$. Calculate the discriminant: $$\Delta = 5^2 - 4 \times 1 \times (-24) = 25 + 96 = 121.$$ So, $$y = \frac{-5 \pm \sqrt{121}}{2} = \frac{-5 \pm 11}{2}.$$ 7. **Find the two possible solutions:** - For the plus sign: $$y = \frac{-5 + 11}{2} = \frac{6}{2} = 3.$$ - For the minus sign: $$y = \frac{-5 - 11}{2} = \frac{-16}{2} = -8.$$ 8. **Check the solutions against the domain and original equation:** - For $y=3$: Check domain: $-1 \leq 3 \leq \frac{25}{3}$ is true. Check original equation: $$\sqrt{25 - 3(3)} = \sqrt{25 - 9} = \sqrt{16} = 4,$$ and $$3 + 1 = 4,$$ so $y=3$ is valid. - For $y=-8$: Check domain: $-1 \leq -8$ is false, so discard $y=-8$. **Final answer:** $$y = 3.$$