Which value for the constant d makes x=-2 an extraneous solution in the following equation? {:[sqrt(6-15 x)=5x+d],[d=]:}

Question

Which value for the constant  d makes  x=-2 an extraneous solution in the following equation?  {:[sqrt(6-15 x)=5x+d],[d=]:}

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Given Equation: We are given the equation sqrt(6-15x) = 5x + d and we need to find the value of d that makes x = -2 an extraneous solution. An extraneous solution is a solution that is derived from an algebraic manipulation but is not a true solution to the original equation. To find d, we will substitute x = -2 into the equation and solve for d.Substitute x = -2: Substitute x = -2 into the left side of the equation sqrt(6-15x). Calculation: sqrt(6-15(-2)) = sqrt(6 + 30) = sqrt(36)Simplify the Square Root: Simplify the square root. Calculation: sqrt(36) = 6Substitute x = -2 (Right Side): Now, substitute x = -2 into the right side of the equation 5x + d. Calculation: 5(-2) + d = -10 + dSet the Two Sides Not Equal: Since x = -2 is supposed to be an extraneous solution, the two sides of the equation should not be equal. Therefore, we set the simplified left side (which is 6) not equal to the simplified right side (-10 + d). Equation: 6 ≠ -10 + dSolve for d: Solve for d by adding 10 to both sides of the inequality. Calculation: 6 + 10 ≠ d 16 ≠ dExtraneous Solution: We have found that for x = -2 to be an extraneous solution, d cannot be equal to 16. Therefore, any value of d other than 16 will make x = -2 an extraneous solution to the equation sqrt(6-15x) = 5x + d.

Which value for the constant $d$ makes $x=-2$ an extraneous solution in the following equation? $\newline$ $\begin{array}{l} \sqrt{6-15 x}=5 x+d \\ d= \end{array}$

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Which value for the constant d d d makes x=−2 x=-2 x=−2 an extraneous solution in the following equation?\newline6−15x=5x+dd= \begin{array}{l} \sqrt{6-15 x}=5 x+d \\ d= \end{array} 6−15x​=5x+dd=​

Which value for the constant $d$ makes $x=-2$ an extraneous solution in the following equation? $\newline$ $\begin{array}{l} \sqrt{6-15 x}=5 x+d \\ d= \end{array}$