Problem 43
Question
Find all real solutions of the equation. \(x^{2}-\sqrt{5} x+1=0\)
Step-by-Step Solution
Verified Answer
The solutions are \(x = \frac{\sqrt{5} + 1}{2}\) and \(x = \frac{\sqrt{5} - 1}{2}\).
1Step 1: Understand the Equation
The equation provided is a quadratic equation: \(x^2 - \sqrt{5}x + 1 = 0\). Our goal is to find the values of \(x\) that satisfy this equation.
2Step 2: Identify the Coefficients
In the quadratic equation \(ax^2 + bx + c = 0\), identify the coefficients: \(a = 1\), \(b = -\sqrt{5}\), and \(c = 1\). These coefficients will be used in the quadratic formula.
3Step 3: Apply the Quadratic Formula
The quadratic formula is given by: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values of \(a\), \(b\), and \(c\) into the formula: \(x = \frac{-(-\sqrt{5}) \pm \sqrt{(-\sqrt{5})^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}\).
4Step 4: Calculate the Discriminant
Calculate the discriminant \(b^2 - 4ac\): \((-\sqrt{5})^2 - 4 \cdot 1 \cdot 1 = 5 - 4 = 1\). The discriminant is 1, which means there are two distinct real solutions.
5Step 5: Find the Roots
Using the quadratic formula, calculate the roots: \(x = \frac{\sqrt{5} \pm \sqrt{1}}{2}\). This simplifies to \(x = \frac{\sqrt{5} \pm 1}{2}\). Therefore, the two solutions are \(x = \frac{\sqrt{5} + 1}{2}\) and \(x = \frac{\sqrt{5} - 1}{2}\).
Key Concepts
Real SolutionsDiscriminantQuadratic FormulaCoefficients
Real Solutions
In mathematics, a real solution to an equation is a solution that is a real number. Real numbers include all the numbers on the number line without any imaginary components. When we solve a quadratic equation like \(x^2 - \sqrt{5}x + 1 = 0\), our focus is to find values of \(x\) that satisfy the equation within the real number system. When the quadratic equation has a discriminant greater than or equal to zero, it ensures that the solutions are real. For our exercise, the discriminant is 1, which clearly indicates that this equation has two real solutions. These solutions are\[ x = \frac{\sqrt{5} + 1}{2} \] and \[ x = \frac{\sqrt{5} - 1}{2} \]. They satisfy the equation without introducing any imaginary numbers.
Discriminant
The discriminant plays a crucial role in determining the nature of the roots of a quadratic equation. The discriminant is part of the quadratic formula, represented as \(b^2 - 4ac\). This expression helps us conclude whether the roots are real or complex:
- If the discriminant is positive, there are two distinct real solutions.
- If it is zero, there is exactly one real solution, sometimes called a repeated or double root.
- If the discriminant is negative, the solutions are complex or imaginary, meaning they do not appear on the real number line.
Quadratic Formula
The quadratic formula is a reliable method to solve any quadratic equation of the form \(ax^2 + bx + c = 0\). It is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula can find the roots of any quadratic equation by simply substituting the values of the coefficients \(a\), \(b\), and \(c\). In our particular exercise, we had the coefficients \(a = 1\), \(b = -\sqrt{5}\), and \(c = 1\). By substituting these into the quadratic formula, the process allows you to:
- Compute the value of \(x\) using \(-b\), making it \(\sqrt{5}\) since \(-(-\sqrt{5}) = \sqrt{5}\).
- Then adding and subtracting the square root of the discriminant, here \(\sqrt{1}\), to get the two solutions.
Coefficients
Coefficients in a quadratic equation are the numbers that multiply each term in the equation \(ax^2 + bx + c = 0\). Identifying these coefficients is the initial step in solving any quadratic equation. In our example, the equation \(x^2 - \sqrt{5}x + 1 = 0\) features the coefficients:
- \(a = 1\)
- \(b = -\sqrt{5}\)
- \(c = 1\)
Other exercises in this chapter
Problem 43
Evaluate the expression and write the result in the form \(a+b i .\) $$ \sqrt{-25} $$
View solution Problem 43
1–54 ? Find all real solutions of the equation. $$ \sqrt{2 x+1}+1=x $$
View solution Problem 43
\(9- 46\) The given equation is either linear or equivalent to a linear equation. Solve the equation. $$ \frac{x}{2 x-4}-2=\frac{1}{x-2} $$
View solution Problem 44
Solve the inequality. Express the answer using interval notation. $$ \frac{1}{|2 x-3|} \leq 5 $$
View solution