Which value for the constant d makes x=-1 an extraneous solution in the following equation? {:[sqrt(8-x)=2x+d],[d=]:}

Question

Which value for the constant  d makes  x=-1 an extraneous solution in the following equation?  {:[sqrt(8-x)=2x+d],[d=]:}

Bytelearn · Accepted Answer

Understanding extraneous solutions: Understand what an extraneous solution is. An extraneous solution is a solution that emerges from the process of solving an equation but is not a valid solution to the original equation. In this case, we want x = -1 to be an extraneous solution, meaning that when we substitute x = -1 into the equation, it should not satisfy the original equation.Substituting x = -1 into the left side: Substitute x = -1 into the left side of the equation. We substitute x = -1 into sqrt(8-x) to see what the left side of the equation would equal if x were -1. Calculation: sqrt(8 - (-1)) = sqrt(8 + 1) = sqrt(9)Simplifying the result: Simplify the result from Step 2. Simplifying sqrt(9) gives us 3, since the square root of 9 is 3. Calculation: sqrt(9) = 3Substituting x = -1 into the right side: Substitute x = -1 into the right side of the equation. We substitute x = -1 into 2x + d to find out what the right side of the equation would equal if x were -1. Calculation: 2(-1) + d = -2 + dSetting the left and right sides equal: Set the results from Step 3 and Step 4 equal to each other. Since we want x = -1 to be an extraneous solution, the left side of the equation (which we found to be 3) should not equal the right side of the equation (which is -2 + d) when x = -1. Equation: 3 ≠ -2 + dSolving for d: Solve for d. To find the value of d that makes x = -1 an extraneous solution, we need to find a value of d that does not satisfy the equation 3 = -2 + d. Calculation: If we were to solve 3 = -2 + d, we would add 2 to both sides to isolate d. 3 + 2 = d 5 = d However, since we want x = -1 to be an extraneous solution, we need to choose a value for d that is not equal to 5.Concluding the value of d: Conclude the value of d. Any value of d other than 5 will make x = -1 an extraneous solution for the equation sqrt(8-x) = 2x + d. Therefore, d can be any real number except 5.

Which value for the constant $d$ makes $x=-1$ an extraneous solution in the following equation? $\newline$ $\begin{array}{l} \sqrt{8-x}=2 x+d \\ d= \end{array}$

Full solution

More problems from Does x satisfy the equation?

Related topics

Which value for the constant d d d makes x=−1 x=-1 x=−1 an extraneous solution in the following equation?\newline8−x=2x+dd= \begin{array}{l} \sqrt{8-x}=2 x+d \\ d= \end{array} 8−x​=2x+dd=​

Which value for the constant $d$ makes $x=-1$ an extraneous solution in the following equation? $\newline$ $\begin{array}{l} \sqrt{8-x}=2 x+d \\ d= \end{array}$