To solve the quadratic equation $-3(x^2 - 1) = 27$ by taking square roots, follow these steps:

Step 1: Expand and simplify

Start by expanding the left-hand side and simplifying the equation: $-3(x^2 - 1) = 27$ Distribute the $-3$ across the terms inside the parentheses: $-3x^2 + 3 = 27$ Now, isolate the $x^2$ term. Subtract 3 from both sides: $-3x^2 = 27 - 3$ $-3x^2 = 24$

Step 2: Divide by $-3$

Next, divide both sides of the equation by $-3$ to solve for $x^2$: $x^2 = \frac{24}{-3}$ $x^2 = -8$

Step 3: Take the square root

To solve for $x$, take the square root of both sides: $x = \pm \sqrt{-8}$ Since $\sqrt{-8}$ involves the square root of a negative number, we express it using imaginary numbers: $x = \pm \sqrt{8}i$

Step 4: Simplify the square root

We can simplify $\sqrt{8}$ as follows: $\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$ Thus, the solutions become: $x = \pm 2\sqrt{2}i$

Final Answer:

The solutions to the equation are: $x = 2\sqrt{2}i \quad \text{or} \quad x = -2\sqrt{2}i$

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Would you like more details on any of the steps, or have further questions? Here are some related questions for exploration:

1. How do you solve a quadratic using the method of completing the square?
2. What are imaginary numbers and how are they used in solving equations?
3. What is the general formula for solving quadratic equations?
4. How do you solve a quadratic equation when the coefficient of $x^2$ is not 1?
5. How do complex solutions arise in quadratic equations?

Tip: When you encounter the square root of a negative number, always use $i$ (the imaginary unit) to express the result!