Which value for the constant d makes y=20 an extraneous solution in the following equation? {:[sqrt((1)/(2)y-1)=(3)/(4)y+d],[d=]:}

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Which value for the constant  d makes  y=20 an extraneous solution in the following equation?  {:[sqrt((1)/(2)y-1)=(3)/(4)y+d],[d=]:}

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Identify equation and condition: Identify the given equation and the condition for an extraneous solution. We are given the equation sqrt(1/2 * y - 1) = (3/4) * y + d and we need to find the value of d that makes y=20 an extraneous solution.Substitute y=20 into equation: Substitute y=20 into the equation to find the value of d that would make this solution extraneous. Substitute y=20 into sqrt(1/2 * y - 1) = (3/4) * y + d. Equation after substitution: sqrt(1/2 * 20 - 1) = (3/4) * 20 + d.Simplify left side of equation: Simplify the left side of the equation. Simplify sqrt(1/2 * 20 - 1) to sqrt(10 - 1) which simplifies to sqrt(9).Simplify right side of equation: Simplify the right side of the equation. Simplify (3/4) * 20 to 15. Now the equation is sqrt(9) = 15 + d.Simplify square root on left side: Simplify the square root on the left side of the equation. sqrt(9) simplifies to 3. Now the equation is 3 = 15 + d.Solve for d: Solve for d. Subtract 15 from both sides of the equation to isolate d. 3 - 15 = d d = -12Verify value of d: Verify if the value of d makes y=20 an extraneous solution. An extraneous solution is one that does not satisfy the original equation. Since we derived d by substituting y=20 into the equation, we need to check if this value of d leads to a contradiction when we substitute y=20 back into the original equation.Substitute y=20 and d=-12 into original equation: Substitute y=20 and d=-12 into the original equation and check for a contradiction. Substitute y=20 and d=-12 into sqrt(1/2 * y - 1) = (3/4) * y + d. Equation after substitution: sqrt(1/2 * 20 - 1) = (3/4) * 20 - 12. Simplify the equation: sqrt(10 - 1) = 15 - 12. Simplify further: sqrt(9) = 3. Since sqrt(9) is indeed 3, there is no contradiction, and y=20 is not an extraneous solution with d=-12. This means we made a mistake in our assumption that d=-12 would make y=20 an extraneous solution.

Which value for the constant $d$ makes $y=20$ an extraneous solution in the following equation? $\newline$ $\begin{array}{l} \sqrt{\frac{1}{2} y-1}=\frac{3}{4} y+d \\ d=\square \end{array}$

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Which value for the constant $d$ makes $y=20$ an extraneous solution in the following equation? $\newline$ $\begin{array}{l} \sqrt{\frac{1}{2} y-1}=\frac{3}{4} y+d \\ d=\square \end{array}$