Chapter 8: Problem 39
Use the quadratic formula to solve each equation. These equations have real solutions and complex but not real solutions. See Examples 1 through 4. $$ (n-2)^{2}=2 n $$
Short Answer
Expert verified
The solutions are \(n = 3 + \sqrt{5}\) and \(n = 3 - \sqrt{5}\).
Step by step solution
01
Expand the Equation
First, expand the left-hand side of the equation. The expression \((n-2)^2\) expands to \(n^2 - 4n + 4\). So the equation becomes: \[ n^2 - 4n + 4 = 2n \]
02
Rearrange the Equation
Next, rearrange the equation to bring all terms to one side, resulting in a standard quadratic form. Subtract \(2n\) from both sides: \[ n^2 - 4n + 4 - 2n = 0 \] Simplify the equation to: \[ n^2 - 6n + 4 = 0 \]
03
Identify Coefficients
Identify the coefficients of the quadratic equation \(n^2 - 6n + 4 = 0\). Here, \(a = 1\), \(b = -6\), and \(c = 4\).
04
Apply the Quadratic Formula
Use the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots of the equation. Substitute \(a = 1\), \(b = -6\), and \(c = 4\) into the formula: \[ n = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \]
05
Simplify Under the Square Root
Calculate the expression under the square root, known as the discriminant. \[ b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot 4 = 36 - 16 = 20 \] So, the discriminant is 20.
06
Simplify the Solutions
Substitute the discriminant back into the quadratic formula and simplify: \[ n = \frac{6 \pm \sqrt{20}}{2} \] Simplifying further, since \(\sqrt{20} = 2\sqrt{5}\): \[ n = \frac{6 \pm 2\sqrt{5}}{2} \] The final solutions are: \[ n = 3 \pm \sqrt{5} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Discriminant
The discriminant is a crucial part of solving quadratic equations. It's the part under the square root in the quadratic formula: \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( b^2 - 4ac \) is the discriminant.
The value of the discriminant determines the nature of the roots of the quadratic equation:
The value of the discriminant determines the nature of the roots of the quadratic equation:
- If it's positive, the equation has two distinct real roots.
- If it's zero, the equation has exactly one real root, known as a double root.
- If it's negative, the equation gives complex solutions with no real roots.
Quadratic Equations
Quadratic equations are polynomial equations of the second degree. They typically look like this: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In our exercise, the standard form of the given equation is \( n^2 - 6n + 4 = 0 \).
To solve them, we use the quadratic formula mentioned earlier. But before that, we often rearrange the original equation into this standard form. A quadratic equation can also be solved by:
To solve them, we use the quadratic formula mentioned earlier. But before that, we often rearrange the original equation into this standard form. A quadratic equation can also be solved by:
- Factoring, when it is possible and easy.
- Completing the square, which is another systematic method.
- Using graphs to estimate solutions.
Complex Solutions
If the discriminant is negative, the equation will result in complex solutions. Complex numbers are formed by a real part and an imaginary part: \( a + bi \); where \( i \) is the imaginary unit with the property \( i^2 = -1 \).
Real solutions exist when the discriminant is zero or positive as in our exercise. In contrast, a negative discriminant results in imaginary solutions, changing the equation outcome significantly. Here’s how it works:
Real solutions exist when the discriminant is zero or positive as in our exercise. In contrast, a negative discriminant results in imaginary solutions, changing the equation outcome significantly. Here’s how it works:
- When roots are complex, you'll see the use of \( i \), showing real parts and imaginary parts, indicating there are no real number solutions.
- These complex solutions are conjugates of each other for any quadratic equation and are represented as \( a + bi \) and \( a - bi \).