Chapter 8: Problem 29
Find \(x\) and \(y\) so that each of the following equations is true. $$\left(x^{2}-2 x\right)+y^{2} i=8+(2 y-1) i$$
Short Answer
Expert verified
The solutions are \(x = 4\) or \(x = -2\), and \(y = 1\).
Step by step solution
01
Equate the real parts
Identify and equate the real parts of the complex numbers. The equation is \(x^2 - 2x = 8\). This is a quadratic equation that we need to solve for \(x\).
02
Solve the quadratic equation
Solve \(x^2 - 2x = 8\) by rearranging it to \(x^2 - 2x - 8 = 0\). Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = -8\).
03
Calculate discriminant
Calculate the discriminant \(b^2 - 4ac = (-2)^2 - 4 \times 1 \times (-8) = 4 + 32 = 36\). The discriminant is positive, indicating two real solutions.
04
Find values of \(x\)
Continuing from the formula for \(x\), \(x = \frac{-(-2) \pm \sqrt{36}}{2 \times 1} = \frac{2 \pm 6}{2}\). So, \(x = 4\) or \(x = -2\).
05
Equate the imaginary parts
Identify and equate the imaginary parts of the complex numbers. The equation is \(y^2 = 2y - 1\). Rearrange this to \(y^2 - 2y + 1 = 0\).
06
Solve for \(y\)
Notice that \(y^2 - 2y + 1 = (y - 1)^2 = 0\). Thus, \(y = 1\). There is only one solution for \(y\).
07
Verify solutions
Insert \(x = 4\) and \(x = -2\) back into the imaginary equation to ensure they can coexist with \(y = 1\). In both cases, the imaginary equation \((1)^2 = 2(1) - 1\) holds true, verifying \(y = 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a fundamental concept in mathematics and represents any equation in the standard form \( ax^2 + bx + c = 0 \). This type of equation is termed 'quadratic' because it involves a squared term, \( x^2 \). Here, \( a \), \( b \), and \( c \) are constants, where \( a \) cannot be zero. In the original exercise, the equation \( x^2 - 2x = 8 \) was identified as a quadratic equation. By rearranging it into its standard form as \( x^2 - 2x - 8 = 0 \), we can apply various methods to solve it.
Common methods to solve a quadratic equation include:
Common methods to solve a quadratic equation include:
- Factoring: Breaking down the equation into factors, if possible.
- Quadratic Formula: Using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the solutions.
- Completing the Square: Rewriting the equation in the form of \( (x+p)^2 = q \).
Discriminant
The discriminant is a vital part of solving quadratic equations. It is derived from the formula \( b^2 - 4ac \), and it tells us about the nature of the roots of the quadratic equation.
Here's what the value of the discriminant indicates:
Here's what the value of the discriminant indicates:
- Positive Discriminant: Two distinct real roots. This was the situation in the exercise where the discriminant was calculated to be 36, leading to two real solutions for \( x \).
- Zero Discriminant: Exactly one real root, often called a repeated root.
- Negative Discriminant: No real roots, but two complex conjugate roots.
Real and Imaginary Parts
In the realm of complex numbers, a number is composed of two parts: the real part and the imaginary part. A complex number is generally expressed in the form \( a + bi \), where \( a \) is the real part and \( b \) is the coefficient of the imaginary part (noting that \( i \) is the imaginary unit with the property \( i^2 = -1 \)).
For the problem at hand, it involved equating both the real and imaginary parts separately:
For the problem at hand, it involved equating both the real and imaginary parts separately:
- Real Part Equations: Aligning the real numbers on both sides of the equation helped to form the quadratic equation \( x^2 - 2x = 8 \). Solving this equation led to the solutions for \( x \).
- Imaginary Part Equations: Similarly, aligning the imaginary parts \( y^2i = (2y - 1)i \), yielded \( y^2 = 2y - 1 \), giving the single solution for \( y \).