Problem 33 - Precalculus

Question

Solve each equation by factoring. $$x^{2}+(2 \sqrt{5}) x+5=0$$

Step-by-Step Solution

Answer

The equation has one repeated root: \( x = -\sqrt{5} \).

Step 1: 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 \( c = 5 \).

Step 2: 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: \( (2\sqrt{5})^2 - 4 \times 1 \times 5 \).

Step 3: 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 \( 20 - 20 = 0 \).

Step 4: 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 \( x = \frac{-(2\sqrt{5}) \pm \sqrt{0}}{2} \).

Step 5: 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 \( x = -\sqrt{5} \).

Key Concepts

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:

• \(a = 1\)
• \(b = 2\sqrt{5}\)
• \(c = 5\)

Understanding that this is a quadratic equation allows you to utilize specific techniques, such as factoring, graphing, completing the square, or using the quadratic formula, to find its roots. Each method will give you the solutions to the equation, which are the values of \(x\) that make the equation true.

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.

For the equation \(x^2 + (2\sqrt{5})x + 5 = 0\), calculating the discriminant involves:

• \(b = 2\sqrt{5}\), so \(b^2 = (2\sqrt{5})^2 = 4 \times 5 = 20\).
• \(4ac = 4 \times 1 \times 5 = 20\).
• Therefore, \(b^2 - 4ac = 20 - 20 = 0\).

The result, zero, indicates a repeated real root.

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 \(x^2 + (2\sqrt{5})x + 5 = 0\):

• \(b = 2\sqrt{5}\), so \(-b = -2\sqrt{5}\).
• \(a = 1\), so \(2a = 2\).
• Thus, the root is \(x = \frac{-2\sqrt{5}}{2} = -\sqrt{5}\).

This repeated root, \(x = -\sqrt{5}\), is the sole solution to the equation, appearing twice on the graph of the quadratic. This particular point is known as the vertex of the parabola formed by the quadratic function.