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Chapter 10: Problem 92

Solve the equation. $$4 v^{2}+9=0$$

Short Answer

Expert verified
The solutions to the equation \(4v^{2} + 9 = 0\) are \(v = 3i/2\) and \(v = -3i/2\).

Step by step solution

01

Isolate the variable term

To start off, we want to isolate the variable term \(4v^{2}\). We can do this by subtracting 9 from both sides of the given equation. This gives us \(4v^{2} = -9\).
02

Remove the coefficient of the variable

The variable \(v\) is being multiplied by a coefficient of 4. We need to remove this coefficient. We do this by dividing the entire equation by 4. This then gives us an updated equation \(v^{2} = -9/4\).
03

Solve for the variable

Now that we only have \(v^{2}\) left, we can solve for \(v\) by taking the square root of both sides. However, to take a square root of a negative number we must introduce imaginary numbers. So \(v = \pm \sqrt{-9/4}\) gives us the solutions \(v = \pm \frac{3i}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Imaginary Numbers
When dealing with quadratic equations like the one given in the exercise, sometimes you might end up having to take the square root of a negative number. This is where imaginary numbers come into play. Imaginary numbers are essential in scenarios where you're unable to find real number solutions. The unit imaginary number is represented by \( i \), which is defined as \( \sqrt{-1} \). With this, any negative number under the square root can be expressed using \( i \).
  • For example, \( \sqrt{-9} \) can be rewritten as \( \sqrt{9} \times \sqrt{-1} = 3i \).
This concept allows us to handle equations that have no real solutions, opening up the realm of complex numbers.
Square Roots
Understanding how to manipulate square roots is crucial in solving quadratic equations, especially those involving imaginary numbers. The square root operation reverses the squaring of a number. When you see an equation like \( v^2 = -9/4 \), the natural step is to find \( v \) by taking the square root of both sides.However, when taking the square root of a negative number, it's essential to apply the imaginary unit \( i \). Thus, \( \sqrt{-9/4} \) becomes \( \sqrt{9/4} \times i = \frac{3i}{2} \).
  • Breaking down: \( \sqrt{9/4} \) turns into \( \frac{\sqrt{9}}{\sqrt{4}} \), resulting in \( \frac{3}{2} \).
  • The negative sign underneath the square root converts to the imaginary unit \( i \), giving \( \frac{3i}{2} \).
This approach allows us to solve previously unsolvable quadratics.
Negative Numbers
Negative numbers are integral to the solution of many quadratic equations. They often appear in situations where direct real solutions don't exist due to their presence under a square root. A negative inside a square root prompts us to use imaginary numbers. In the example of \( 4v^{2} = -9 \), the negative number \(-9\) leads us down the path of incorporating imaginary numbers, specifically converting to \( 3i \) for calculation.
  • In standard operations, resolving such negatives typically involves rearranging the equation.
  • This may include isolating variables and addressing the coefficient of the variable before diving into square root operations.
  • Negative numbers, when isolated properly, pave the way to utilize complex numbers for solution finding.
Learning to work with negative numbers effectively is crucial in successfully solving complex quadratic equations.

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