Chapter 1: Problem 33
Solve each equation by factoring. $$x^{2}+(2 \sqrt{5}) x+5=0$$
Short Answer
Expert verified
The equation has one repeated root: \( x = -\sqrt{5} \).
Step by step solution
01
Identify the Equation Type
Notice that the given equation is a quadratic equation in the standard form, which is represented as \( ax^2 + bx + c = 0 \). In this case, \( a = 1 \), \( b = 2 \sqrt{5} \), and \( c = 5 \).
02
Check the Factoring Requirement
We need to determine whether the quadratic can be factored. First, check for factorability by calculating the discriminant \( b^2 - 4ac \). Substitute the values: \( (2\sqrt{5})^2 - 4 \times 1 \times 5 \).
03
Calculate the Discriminant
Calculate \( b^2 - 4ac \) = \( (2\sqrt{5})^2 - 4\cdot1\cdot5\). Calculate each component: \((2\sqrt{5})^2 = 4 \times 5 = 20\) and \(4 \times 1 \times 5 = 20\). Thus, the discriminant is \( 20 - 20 = 0 \).
04
Solve by Using the Discriminant
Since the discriminant is zero, we know the quadratic has a repeated real root. To find this root, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting the values, we have \( x = \frac{-(2\sqrt{5}) \pm \sqrt{0}}{2} \).
05
Finalize the Solution
Since \( \sqrt{0} = 0 \), the solution simplifies to \( x = \frac{-(2\sqrt{5})}{2} \). Simplifying further, \( x = -\sqrt{5} \). Thus, the equation has one root: a repeated value \( x = -\sqrt{5} \).
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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 polynomial equation of degree two, which means it includes an \(x^2\) term as its highest order. The standard form of a quadratic equation is expressed as \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants with \(a eq 0\). The equation you are working with, \(x^2 + (2\sqrt{5})x + 5 = 0\), fits this form where:
- \(a = 1\)
- \(b = 2\sqrt{5}\)
- \(c = 5\)
Discriminant
The discriminant is a key component when solving quadratic equations and helps determine the nature of the roots of a quadratic equation. Given by the formula \(b^2 - 4ac\), it originates from the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Depending on its value, the discriminant informs us:
- If it is positive, there are two distinct real roots.
- If it is zero, there is one repeated real root.
- If it is negative, there are two complex conjugate roots.
- \(b = 2\sqrt{5}\), so \(b^2 = (2\sqrt{5})^2 = 4 \times 5 = 20\).
- \(4ac = 4 \times 1 \times 5 = 20\).
- Therefore, \(b^2 - 4ac = 20 - 20 = 0\).
Repeated Root
When the discriminant of a quadratic equation is zero, it indicates a repeated root. This means the quadratic equation has only one unique solution, although it occurs twice (hence repeated).
Repeated roots occur when a quadratic's graph touches the x-axis at exactly one point without crossing it.
In the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the presence of a zero discriminant simplifies it to \(x = \frac{-b}{2a}\) since \(\pm \sqrt{0} = 0\).
Using this simplified form with the values from the equation \(x^2 + (2\sqrt{5})x + 5 = 0\):
Repeated roots occur when a quadratic's graph touches the x-axis at exactly one point without crossing it.
In the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the presence of a zero discriminant simplifies it to \(x = \frac{-b}{2a}\) since \(\pm \sqrt{0} = 0\).
Using this simplified form with the values from the equation \(x^2 + (2\sqrt{5})x + 5 = 0\):
- \(b = 2\sqrt{5}\), so \(-b = -2\sqrt{5}\).
- \(a = 1\), so \(2a = 2\).
- Thus, the root is \(x = \frac{-2\sqrt{5}}{2} = -\sqrt{5}\).