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Which value for the constant c makes z=4 an extraneous solution in the following equation? {:[sqrt((3)/(2)z+10)=cz+8],[c=]:}

Which value for the constant c c makes z=4 z=4 an extraneous solution in the following equation?\newline32z+10=cz+8c=□ \begin{array}{l} \sqrt{\frac{3}{2} z+10}=c z+8 \\ c=\square \end{array}

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Q. Which value for the constant c c makes z=4 z=4 an extraneous solution in the following equation?\newline32z+10=cz+8c=□ \begin{array}{l} \sqrt{\frac{3}{2} z+10}=c z+8 \\ c=\square \end{array}
  1. Substituting z=4z=4: Let's first substitute z=4z=4 into the left side of the equation to see what value we get.\newline(32)z+10=(32)⋅4+10\sqrt{\left(\frac{3}{2}\right)z + 10} = \sqrt{\left(\frac{3}{2}\right)\cdot 4 + 10}
  2. Calculating the value inside the square root: Now, let's calculate the value inside the square root.\newline(32)⋅4+10=6+10=16(\frac{3}{2})\cdot 4 + 10 = 6 + 10 = 16
  3. Taking the square root of 1616: Next, we take the square root of 1616.\newline16=4\sqrt{16} = 4
  4. Substituting z=4z=4 into the right side: Now, let's substitute z=4z=4 into the right side of the equation.cz+8=c⋅4+8cz + 8 = c\cdot4 + 8
  5. Setting up the inequality: Since z=4z=4 is an extraneous solution, the right side of the equation should not equal 44 when z=4z=4. Therefore, we set up the equation c⋅4+8≠4c\cdot 4 + 8 \neq 4.
  6. Isolating cc in the inequality: We solve for cc by isolating it on one side of the inequality.\newlinec×4≠4−8c \times 4 \neq 4 - 8\newlinec×4≠−4c \times 4 \neq -4
  7. Solving for c: Divide both sides by 44 to solve for c. \newlinec≠−44c \neq \frac{-4}{4}\newlinec≠−1c \neq -1

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