Chapter 1: Problem 56
Find the real or imaginary solutions to each equation by using the quadratic formula. $$x^{2}-4 x+9=0$$
Short Answer
Expert verified
The solutions are \(x = 2 + \sqrt{5}i\) and \(x = 2 - \sqrt{5}i\).
Step by step solution
01
- Identify coefficients
For the quadratic equation in the form of \(ax^2 + bx + c = 0\), identify the coefficients: \(a = 1\), \(b = -4\), and \(c = 9\).
02
- Apply the quadratic formula
The quadratic formula is given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Substitute \(a = 1\), \(b = -4\), and \(c = 9\) into the formula: \[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}\].
03
- Simplify inside the square root
Simplify the expression inside the square root: \((-4)^2 = 16\), and \(4 \cdot 1 \cdot 9 = 36\). So, \[b^2 - 4ac = 16 - 36 = -20\].
04
- Calculate the square root of a negative number
Since the value under the square root is negative (\(-20\)), it indicates the presence of imaginary solutions. \(\sqrt{-20} = \sqrt{20}i = 2\sqrt{5}i\).
05
- Complete the quadratic formula
Substitute \(\sqrt{-20} = 2\sqrt{5}i\) back into the formula: \[x = \frac{4 \pm 2\sqrt{5}i}{2}\]. Simplify the fraction: \[x = 2 \pm \sqrt{5}i\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation Solutions
To solve any quadratic equation of the form \(ax^2 + bx + c = 0\), we can use the quadratic formula. This remarkable formula provides the solutions (or roots) for the given quadratic equation.
Here's the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
Let's use this process on the equation: \(x^2 - 4x + 9 = 0\). Here, \(a = 1\), \(b = -4\), and \(c = 9\). Using the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} \].
Simplifying inside the square root, we find \(16 - 36 = -20\). This tells us more about the kinds of solutions we have.
Here's the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
- Identify the coefficients \(a\), \(b\), and \(c\). They are constants in the equation's format.
- Substitute these coefficients into the formula.
- Simplify the equation to find the values of \(x\).
Let's use this process on the equation: \(x^2 - 4x + 9 = 0\). Here, \(a = 1\), \(b = -4\), and \(c = 9\). Using the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} \].
Simplifying inside the square root, we find \(16 - 36 = -20\). This tells us more about the kinds of solutions we have.
Imaginary Numbers
An imaginary number arises when we attempt to take the square root of a negative number. Typically, the square root of a negative number isn't a real number. Instead, it's termed imaginary.
In our formula, the \(\sqrt{-20}\) presents an issue since \(20\) is negative. By convention, we use the imaginary unit \(i\), where \(i = \sqrt{-1}\).
So \(\sqrt{-20} = \sqrt{20}i\). We can further simplify \((\sqrt{20})\) as \(2\sqrt{5}\) because \(20 = 4 \times 5\), and \(\sqrt{4} = 2\). Therefore, \[ \sqrt{-20} = 2\sqrt{5}i \].
Imaginary numbers help us handle roots of negative quantities intuitively.
In our formula, the \(\sqrt{-20}\) presents an issue since \(20\) is negative. By convention, we use the imaginary unit \(i\), where \(i = \sqrt{-1}\).
So \(\sqrt{-20} = \sqrt{20}i\). We can further simplify \((\sqrt{20})\) as \(2\sqrt{5}\) because \(20 = 4 \times 5\), and \(\sqrt{4} = 2\). Therefore, \[ \sqrt{-20} = 2\sqrt{5}i \].
Imaginary numbers help us handle roots of negative quantities intuitively.
Complex Solutions
Complex solutions emerge when the quadratic equation doesn't intersect the x-axis. This happens when the discriminant (\b^2 - 4ac) is negative.
Complex numbers are formed from a real part and an imaginary part. In our equation: \[ x = \frac{4 \pm 2 \sqrt{5}i}{2} \].
Simplifying, we get: \[ x = 2 \pm \sqrt{5}i \].
Here, \(x\) has both a real part \(2\) and an imaginary part \(\pm \sqrt{5}i\). Complex solutions are written as \[ a + bi \] where \(a\) (real part) is \(2\) and \(bi\) (imaginary part) is \(\sqrt{5}i\).
Complex solutions play a crucial role in various fields, including engineering and physics, by simplifying calculations involving oscillations and waves.
So, the final solutions \(x = 2 \pm \sqrt{5}i\) elucidate both the real and imaginary aspects of the complex solutions.
Complex numbers are formed from a real part and an imaginary part. In our equation: \[ x = \frac{4 \pm 2 \sqrt{5}i}{2} \].
Simplifying, we get: \[ x = 2 \pm \sqrt{5}i \].
Here, \(x\) has both a real part \(2\) and an imaginary part \(\pm \sqrt{5}i\). Complex solutions are written as \[ a + bi \] where \(a\) (real part) is \(2\) and \(bi\) (imaginary part) is \(\sqrt{5}i\).
Complex solutions play a crucial role in various fields, including engineering and physics, by simplifying calculations involving oscillations and waves.
So, the final solutions \(x = 2 \pm \sqrt{5}i\) elucidate both the real and imaginary aspects of the complex solutions.