9) \( 15-2(3 y-2)^{\frac{1}{2}}=y-4 \)
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To solve the equation \( 15-2(3 y-2)^{\frac{1}{2}}=y-4 \), let's first isolate the square root. We can rearrange it: 1. Add \( 2(3y - 2)^{\frac{1}{2}} \) to both sides: \( 15 = y - 4 + 2(3y - 2)^{\frac{1}{2}} \) 2. Simplifying, we have: \( y + 1 = 2(3y - 2)^{\frac{1}{2}} \) 3. Divide by 2: \( (y + 1)/2 = (3y - 2)^{\frac{1}{2}} \) 4. Square both sides to eliminate the square root: \( \left( \frac{y + 1}{2} \right)^2 = 3y - 2 \) 5. This gives us: \( \frac{(y + 1)^2}{4} = 3y - 2 \) 6. Multiplying through by 4 to eliminate the fraction: \( (y + 1)^2 = 12y - 8 \) 7. Expanding the left side: \( y^2 + 2y + 1 = 12y - 8 \) 8. Rearranging everything to one side gives: \( y^2 - 10y + 9 = 0 \) Now we can factor this quadratic: 9. Factoring yields: \( (y - 1)(y - 9) = 0 \) 10. Setting each factor to zero gives us: \( y = 1 \) or \( y = 9 \) Lastly, we should check these solutions in the original equation to ensure they don't produce extraneous solutions. When substituting \( y = 1 \) and \( y = 9 \) back into the original equation, both validate, confirming that the solutions are correct: \( y = 1 \) and \( y = 9 \).