Chapter 9: Problem 1
Solve the following equations: (a) \(x^{2}+1=0\) (b) \(x^{2}+4=0\) (c) \(3 x^{2}+7=0\) (d) \(x^{2}+x+1=0\) (e) \(\frac{x^{2}}{2}-x+2=0\) (f) \(-x^{2}-3 x-4=0\) (g) \(2 x^{2}+3 x+3=0\) (h) \(x^{2}+3 x+4=0\)
Short Answer
Expert verified
(a) \(x = \pm i\), (b) \(x = \pm 2i\), (c) \(x = \pm \frac{\sqrt{21}}{3}i\), (d) \(x = \frac{-1 \pm i\sqrt{3}}{2}\), (e) \(x = 1 \pm i\sqrt{3}\), (f) \(x = \frac{3 \pm i\sqrt{7}}{2}\), (g) \(x = \frac{-3 \pm i\sqrt{15}}{4}\), (h) \(x = \frac{-3 \pm i\sqrt{7}}{2}\).
Step by step solution
01
Solve Equation (a)
Equation: \(x^2 + 1 = 0\). Move the constant to the other side: \(x^2 = -1\). To solve for \(x\), take the square root of both sides, giving \(x = \pm \sqrt{-1}\). Since \(\sqrt{-1}\) is an imaginary number, \(x = \pm i\).
02
Solve Equation (b)
Equation: \(x^2 + 4 = 0\). Rearrange to \(x^2 = -4\). Take the square root: \(x = \pm \sqrt{-4}\). This simplifies to \(x = \pm 2i\), as \(\sqrt{-1} = i\) and \(\sqrt{4} = 2\).
03
Solve Equation (c)
Equation: \(3x^2 + 7 = 0\). Rearrange to \(3x^2 = -7\). Divide by 3: \(x^2 = -\frac{7}{3}\). Take the square root: \(x = \pm \sqrt{-\frac{7}{3}}\). The solutions are \(x = \pm \frac{\sqrt{21}}{3}i\).
04
Solve Equation (d) using the Quadratic Formula
Equation: \(x^2 + x + 1 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 1\), and \(c = 1\). The discriminant \(b^2 - 4ac = 1 - 4 = -3\). Thus, \(x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}\).
05
Solve Equation (e) Using the Quadratic Formula
Equation: \(\frac{x^2}{2} - x + 2 = 0\). Rearrange coefficients: \(a = \frac{1}{2}\), \(b = -1\), and \(c = 2\). The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Calculate the discriminant: \(b^2 - 4ac = 1 - 4 \times \frac{1}{2} \times 2 = -3\). Apply the formula: \(x = \frac{1 \pm i\sqrt{3}}{1}\). The solutions are \(x = 1 \pm i\sqrt{3}\).
06
Solve Equation (f) Using the Quadratic Formula
Equation: \(-x^2 - 3x - 4 = 0\). Identify coefficients: \(a = -1\), \(b = -3\), \(c = -4\). Calculate the discriminant: \(b^2 - 4ac = 9 - 16 = -7\). Apply the quadratic formula: \(x = \frac{3 \pm i\sqrt{7}}{2}\). The solutions are \(x = \frac{3 \pm i\sqrt{7}}{2}\).
07
Solve Equation (g) Using the Quadratic Formula
Equation: \(2x^2 + 3x + 3 = 0\). Identify coefficients: \(a = 2\), \(b = 3\), \(c = 3\). Calculate the discriminant \(b^2 - 4ac = 9 - 24 = -15\). Apply the formula: \(x = \frac{-3 \pm i\sqrt{15}}{4}\). The solutions are \(x = \frac{-3 \pm i\sqrt{15}}{4}\).
08
Solve Equation (h) Using the Quadratic Formula
Equation: \(x^2 + 3x + 4 = 0\). Use the quadratic formula with coefficients \(a = 1\), \(b = 3\), \(c = 4\). The discriminant is \(b^2 - 4ac = 9 - 16 = -7\). Thus, \(x = \frac{-3 \pm i\sqrt{7}}{2}\), leading to solutions \(x = \frac{-3 \pm i\sqrt{7}}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations are a central concept in algebra. Each equation takes the general form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable we are solving for. The key characteristic of these equations is that the highest power of the variable \( x \) is 2. Quadratics can be represented as a parabola on a graph, and they often have two solutions, or roots.
It's essential to understand how these roots can be real, repeated, or complex depending on the discriminant, \( b^2 - 4ac \):
It's essential to understand how these roots can be real, repeated, or complex depending on the discriminant, \( b^2 - 4ac \):
- If \( b^2 - 4ac > 0 \), the equation has two distinct real roots.
- If \( b^2 - 4ac = 0 \), there is one real and repeated root.
- If \( b^2 - 4ac < 0 \), the roots are complex, involving imaginary numbers.
Imaginary Numbers
Imaginary numbers come into play when we attempt to take the square root of a negative number—something not possible within the realm of real numbers. The most basic imaginary number is \( i \), which is defined by the property \( i^2 = -1 \). This introduction of \( i \) allows us to handle square roots of negative numbers, which is why they appear frequently in solutions of quadratic equations with negative discriminants.
For example, when solving the equation \( x^2 + 1 = 0 \), rearranging gives \( x^2 = -1 \). The solutions to this are \( x = \pm i \). Simply put:
For example, when solving the equation \( x^2 + 1 = 0 \), rearranging gives \( x^2 = -1 \). The solutions to this are \( x = \pm i \). Simply put:
- The square root of -1 is \( i \).
- If you have \( \sqrt{-b} \), it equals \( i\sqrt{b} \).
Quadratic Formula
The quadratic formula is a tool for finding the roots of any quadratic equation, no matter its nature. Written as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula allows us to directly calculate the roots using the coefficients \( a \), \( b \), and \( c \). To use the quadratic formula effectively:
As seen in the given problems, employing the quadratic formula is part of the standard approach to solving a wide variety of quadratic equations, including those with complex numbers. This formula not only simplifies finding roots, but also enhances your understanding of the structure and solutions of quadratic equations.
- Identify \( a \), \( b \), and \( c \) from the equation \( ax^2 + bx + c = 0 \).
- Calculate the discriminant \( b^2 - 4ac \).
- Plug these values into the quadratic formula to find the roots.
As seen in the given problems, employing the quadratic formula is part of the standard approach to solving a wide variety of quadratic equations, including those with complex numbers. This formula not only simplifies finding roots, but also enhances your understanding of the structure and solutions of quadratic equations.