Problem 51 - Applied Calculus

Question

Solve each equation by factoring or the Quadratic Formula, as appropriate. $$ 3 x^{2}+12=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = 2i \) and \( x = -2i \).

Step 1: Move 12 to Right Side

Start by moving the constant term 12 to the right side of the equation, so subtract 12 from both sides to isolate the quadratic expression.The equation becomes:\[ 3x^2 = -12 \]

Step 2: Divide by 3

To simplify the equation, divide every term by 3 to get the coefficient of \( x^2 \) to be 1.This gives:\[ x^2 = -4 \]

Step 3: Recognize a Quadratic Equation

Recognize that the equation \( x^2 = -4 \) is a basic quadratic equation that can be solved using the square root method. The negative sign indicates imaginary solutions.

Step 4: Solve for x

Take the square root of both sides of the equation. Remember to include both the positive and negative roots and consider that the square root of a negative number involves the imaginary unit \( i \).\[ x = \sqrt{-4} \] \[ x = \pm 2i \]

Key Concepts

Factoring

Factoring is one of the techniques used to solve quadratic equations. It involves expressing the quadratic equation as a product of two binomials. Factoring is most effective when the quadratic can be easily rewritten into products of simpler expressions. This means the equation usually has real roots.

• Look for two numbers that multiply to give you the constant term (the number without a variable) and add or subtract to get the middle coefficient.
• Once you identify these factors, rewrite the quadratic equation using these numbers.
• Finally, apply the zero-product property: if the product of these factors is zero, then one of the factors must be zero.

This method won't work directly in scenarios where complex or imaginary numbers come into play, like in our exercise, because perfect square factors do not exist in real numbers for negative constants.

Quadratic Formula

The Quadratic Formula is another powerful tool to solve quadratic equations when factoring is not straightforward. It applies to any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Notice the "+/-" sign, which means that this formula will always produce two solutions.

• "\( b^2 - 4ac \)" is called the discriminant and determines the nature of the roots.
• If positive, real and distinct solutions exist. If zero, real and repeated solutions. If negative, solutions are complex.

In the case of our exercise, using the quadratic formula would involve recognizing that the discriminant \((0 - 4(3)(-12))\) is negative, thus indicating complex roots similar to the result obtained with the square root method.

Imaginary Numbers

Imaginary numbers come into play when we take square roots of negative numbers. The basic imaginary unit is \( i \), where \( i^2 = -1 \). For quads with negative discriminants, imaginary numbers ensure equations are solvable.

• When confronted with \( \sqrt{-a} \), where \( a > 0 \), the result becomes \( i\sqrt{a} \).
• Imaginary numbers are part of complex numbers, which can be expressed in \( a + bi \) form, where \( a \) and \( b \) are real numbers.

In the exercise, the solution was \( x = \pm 2i \). It demonstrates the necessity of imaginary numbers when dealing with negative expressions inside a square root, highlighting their fundamental role in expanding the solution set for quadratic equations.

Square Root Method

The Square Root Method is often used for simple quadratic equations when the equation is perfectly set up for taking square roots directly. It's particularly useful when the quadratic term has no linear component or when it has been factored or simplified into a square.

• First, ensure the quadratic term is isolated on one side of the equation, such as \( x^2 = -4 \).
• The next step is to take the square root of both sides. Remember, taking the square root yields both a positive and negative result.
• For negative numbers, include the imaginary unit \( i \), since you're essentially taking a root of a negative value.

In our exercise, applying this method was advantageous because it quickly resulted in \( x = \pm 2i \) without needing complex algebraic manipulation, demonstrating its simplicity especially when dealing with imaginary or negative-valued square roots.