Which value for the constant c makes z=2 an extraneous solution in the following equation? {:[sqrt(7-3z)=3+cz],[c=]:}

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Which value for the constant  c makes  z=2 an extraneous solution in the following equation?  {:[sqrt(7-3z)=3+cz],[c=]:}

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Given Equation: We are given the equation sqrt(7-3z) = 3 + cz and we need to find the value of c that makes z=2 an extraneous solution. An extraneous solution is a solution that is derived from an algebraic manipulation but is not a solution to the original equation. To find the value of c, we will substitute z=2 into the equation and solve for c.Substitute z=2: Substitute z=2 into the equation sqrt(7-3z) = 3 + cz. Equation becomes: sqrt(7-3(2)) = 3 + c(2)Simplify the Equation: Simplify the equation by performing the operation inside the square root and the multiplication by c. Equation becomes: sqrt(7-6) = 3 + 2c Which simplifies to: sqrt(1) = 3 + 2cIsolate c: Since sqrt(1) is 1, the equation simplifies to: 1 = 3 + 2cDivide by 2: To solve for c, we need to isolate c on one side of the equation. We do this by subtracting 3 from both sides of the equation. 1 - 3 = 3 + 2c - 3 Which simplifies to: -2 = 2cVerify Extraneous Solution: Now, divide both sides of the equation by 2 to solve for c. -2 / 2 = 2c / 2 Which simplifies to: -1 = cSubstitute c=-1 and z=2: We have found that c must be -1 for z=2 to be an extraneous solution to the equation sqrt(7-3z) = 3 + cz. However, we need to check if z=2 is indeed an extraneous solution by substituting c=-1 and z=2 back into the original equation and verifying if it does not hold true.Simplify Right Side: Substitute c=-1 and z=2 into the original equation sqrt(7-3z) = 3 + cz. Equation becomes: sqrt(7-3(2)) = 3 + (-1)(2) Which simplifies to: sqrt(1) = 3 - 2Verify Equation: Simplify the right side of the equation: sqrt(1) = 1 Since sqrt(1) is also 1, the equation holds true, which means that z=2 is not an extraneous solution for c=-1. This contradicts our initial assumption, indicating a mistake in our reasoning. We were supposed to find a value of c that makes z=2 an extraneous solution, but instead, we found a value of c that makes z=2 a valid solution. Therefore, there is a math error in our reasoning process.

Which value for the constant $c$ makes $z=2$ an extraneous solution in the following equation? $\newline$ $\begin{array}{l} \sqrt{7-3 z}=3+c z \\ c= \end{array}$

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Which value for the constant c c c makes z=2 z=2 z=2 an extraneous solution in the following equation?\newline7−3z=3+czc= \begin{array}{l} \sqrt{7-3 z}=3+c z \\ c= \end{array} 7−3z​=3+czc=​

Which value for the constant $c$ makes $z=2$ an extraneous solution in the following equation? $\newline$ $\begin{array}{l} \sqrt{7-3 z}=3+c z \\ c= \end{array}$