Chapter 9: Problem 51
Solve by extracting the roots and then solve by using the quadratic formula. Check answer. $$ 4 y 2+3=0 $$
Short Answer
Expert verified
The solutions are: \(y = \pm \frac{i \sqrt{3}}{2}\).
Step by step solution
01
Identify Equation Type
The given equation is a quadratic equation, which can be written as \(4y^2 + 3 = 0\). This is in the form \(ay^2 + by + c = 0\), where \(a = 4\), \(b = 0\), and \(c = 3\).
02
Solve by Extracting Roots
First, isolate the quadratic term by subtracting 3 from both sides: \(4y^2 = -3\). Next, divide both sides by 4: \(y^2 = -\frac{3}{4}\). Since there is a negative sign under the square root, we introduce the imaginary unit \(i\): \(y = \pm \sqrt{-\frac{3}{4}}\), which simplifies to \(y = \pm \frac{i \sqrt{3}}{2}\).
03
Solve by Quadratic Formula
Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute \(a = 4\), \(b = 0\), and \(c = 3\) into the formula. \(y = \frac{-0 \pm \sqrt{0^2 - 4 \times 4 \times 3}}{2 \times 4}\). Simplify under the square root: \(y = \frac{\pm \sqrt{-48}}{8}\). Further simplify: \(y = \frac{\pm i \sqrt{48}}{8}\). Break down \(\sqrt{48}\) into \(4\sqrt{3}\), giving \(y = \frac{\pm 4i \sqrt{3}}{8}\) or \(y = \pm \frac{i \sqrt{3}}{2}\).
04
Verify Solutions
Both methods resulted in \(y = \pm \frac{i \sqrt{3}}{2}\), confirming the solution is consistent and correct. The imaginary unit indicates these are non-real solutions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solving Quadratic Equations
Quadratic equations can appear in many forms but they all share a common pattern: \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. To solve these equations, there are different methods, each suitable for particular scenarios, such as:
- Factoring: This involves expressing the quadratic as a product of two binomials. It works well when the roots are rational numbers.
- Completing the square: A method where you rearrange the equation to form a perfect square trinomial. It’s useful for deriving the quadratic formula itself.
- Quadratic formula: A universal tool that can solve any quadratic equation, regardless of the nature of its roots.
- Extracting roots: When the equation is in the form \( ax^2 + c = 0 \), simplify to \( x^2 = k \) and solve \( x = \pm \sqrt{k} \).
Complex Solutions
When dealing with quadratic equations, sometimes the solutions won't be real numbers. This occurs when the discriminant \((b^2 - 4ac)\) is negative, leading us into the realm of complex numbers. Complex numbers have two components: a real part and an imaginary part, where the imaginary unit \(i\) satisfies \(i^2 = -1\).
For example, in solving \(4y^2 + 3 = 0\), after reducing to \(y^2 = -\frac{3}{4}\), we're faced with taking a square root of a negative number, indicating complex solutions.
For example, in solving \(4y^2 + 3 = 0\), after reducing to \(y^2 = -\frac{3}{4}\), we're faced with taking a square root of a negative number, indicating complex solutions.
- Imaginary Unit \(i\): Represents the square root of -1. Hence, \( \sqrt{-1} = i \).
- Expressing Complex Solutions: In the form \(y = \pm \frac{i \sqrt{3}}{2}\), indicating two imaginary solutions.
- Understanding Negative Discriminants: If \(b^2 - 4ac < 0\), expect solutions in terms of \(i\).
Quadratic Formula
The quadratic formula is a key tool for solving any quadratic equation. It states that for \( ax^2 + bx + c = 0 \), the solutions for \( x \) are given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This powerful formula allows you to find the roots by direct substitution and computation. In our exercise, where \(a = 4\), \(b = 0\), and \(c = 3\), substituting directly into the formula reveals how to handle equations with complex roots:
- Substitute Values: Replace \(a, b, \text{ and } c \) into the formula.
- Compute the Discriminant: Find \(b^2 - 4ac\). If negative, indicates complex roots.
- Solve for \(x\): Complete the computation, watching for negative discriminants which require \(i\).