Chapter 8: Problem 9
Solve. (Find all complex-number solutions.) \(u^{2}+2 u-4=0\)
Short Answer
Expert verified
The solutions are \( u = -1 + \sqrt{5}\) and \(u = -1 - \sqrt{5}\).
Step by step solution
01
Identify coefficients
The given quadratic equation is in the form \(u^2 + 2u - 4 = 0\). Here, identify the coefficients:\(a = 1\), \(b = 2\), and \(c = -4\).
02
Use the quadratic formula
The quadratic formula is defined as: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the coefficients \(a, b\), and \(c\) into the quadratic formula.
03
Calculate the discriminant
Calculate the discriminant \(\Delta = b^2 - 4ac\). Substitute the values: \(\Delta = 2^2 - 4 \cdot 1 \cdot (-4) = 4 + 16 = 20\).
04
Compute the solutions
Substitute \(\Delta\) back into the quadratic formula: \(u = \frac{-2 \pm \sqrt{20}}{2 \cdot 1}\). Simplify the square root of 20 (\(\sqrt{20} = 2\sqrt{5}\)) and solve for \(u\): \(u = \frac{-2 \pm 2\sqrt{5}}{2}\).
05
Simplify the solutions
Simplify the fractions: \(u = -1 \pm \sqrt{5}\). So, the two solutions are \(u = -1 + \sqrt{5}\) and \(u = -1 - \sqrt{5}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). The quadratic formula provides a way to find the solutions to such equations. The formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This helps you avoid the tedious process of factoring, especially when factors aren’t obvious. The symbol \(\pm\) means you will get two solutions: one with addition and one with subtraction.
When solving the equation \(u^2 + 2u - 4 = 0\), we identify coefficients: \(a = 1\), \(b = 2\), and \(c = -4\). We plug these into the quadratic formula as follows:
\[ u = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
Next, we simplify under the square root, leading to the discriminant.
When solving the equation \(u^2 + 2u - 4 = 0\), we identify coefficients: \(a = 1\), \(b = 2\), and \(c = -4\). We plug these into the quadratic formula as follows:
\[ u = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
Next, we simplify under the square root, leading to the discriminant.
Discriminant
The discriminant is a specific part of the quadratic formula found inside the square root: \(b^2 - 4ac\). It tells us about the nature of the roots of the quadratic equation.
For the equation \(u^2 + 2u - 4 = 0\), we calculate: \[ \Delta = 2^2 - 4 \cdot 1 \cdot (-4) = 4 + 16 = 20 \]
Since \(\Delta > 0\), it indicates that there are two distinct real roots. If \(\Delta = 0\), there is exactly one real root. If \(\Delta < 0\), there are two complex roots. However, we have real roots here due to the positive discriminant.
Substituting the discriminant back into the formula gives us:
\[ u = \frac{-2 \pm \sqrt{20}}{2 \cdot 1} \]
For the equation \(u^2 + 2u - 4 = 0\), we calculate: \[ \Delta = 2^2 - 4 \cdot 1 \cdot (-4) = 4 + 16 = 20 \]
Since \(\Delta > 0\), it indicates that there are two distinct real roots. If \(\Delta = 0\), there is exactly one real root. If \(\Delta < 0\), there are two complex roots. However, we have real roots here due to the positive discriminant.
Substituting the discriminant back into the formula gives us:
\[ u = \frac{-2 \pm \sqrt{20}}{2 \cdot 1} \]
Complex-Number Solutions
Even though the discriminant for our example is positive, let's discuss what happens when a quadratic equation has complex solutions. Complex roots occur when the discriminant \(\Delta\) is negative. In such cases, the square root of a negative number is not real but complex.
To illustrate, let’s assume \(\Delta = -4\). You will insert this into the quadratic formula:
\[ x = \frac{-b \pm \sqrt{-4}}{2a} \]
Here, \(\sqrt{-4} = 2i\) where \(i\) is the imaginary unit with \(i^2 = -1\). Hence the solutions would be in the form:
\[ x = \frac{-b \pm 2i}{2a} \]
Resulting in complex number solutions involving \(i\). Complex numbers have real and imaginary components and are written as \(a + bi\). This approach is useful when dealing with quadratic equations with negative discriminants.
To illustrate, let’s assume \(\Delta = -4\). You will insert this into the quadratic formula:
\[ x = \frac{-b \pm \sqrt{-4}}{2a} \]
Here, \(\sqrt{-4} = 2i\) where \(i\) is the imaginary unit with \(i^2 = -1\). Hence the solutions would be in the form:
\[ x = \frac{-b \pm 2i}{2a} \]
Resulting in complex number solutions involving \(i\). Complex numbers have real and imaginary components and are written as \(a + bi\). This approach is useful when dealing with quadratic equations with negative discriminants.