Question

8.6 solve radical equations (homework)
score: 5/16 answered: 5/16
question 6
find all real solutions:
$sqrt{6 - 6c}-9 = 0$
$c=$. (if no solution exists, enter dne)

Explanation:

Step1: Isolate the radical term

Add 9 to both sides of the equation $\sqrt{6 - 6c}-9 = 0$. We get $\sqrt{6 - 6c}=9$.

Step2: Square both sides

Squaring both sides to eliminate the square - root, we have $(\sqrt{6 - 6c})^2=9^2$, which simplifies to $6 - 6c = 81$.

Step3: Solve for c

First, subtract 6 from both sides: $-6c=81 - 6$, so $-6c = 75$. Then divide both sides by - 6, we get $c=-\frac{75}{6}=-\frac{25}{2}$.

Step4: Check the solution

Substitute $c =-\frac{25}{2}$ into the original equation. The left - hand side is $\sqrt{6-6\times(-\frac{25}{2})}-9=\sqrt{6 + 75}-9=\sqrt{81}-9=9 - 9=0$, which is equal to the right - hand side.

Answer:

$-\frac{25}{2}$