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Chapter 1: Problem 64

Find all real solutions of the quadratic equation. $$x^{2}-\sqrt{5} x+1=0$$

Short Answer

Expert verified
The real solutions are \(x_1 = \frac{\sqrt{5} + 1}{2}\) and \(x_2 = \frac{\sqrt{5} - 1}{2}\).

Step by step solution

01

Identify the quadratic equation

The given quadratic equation is \(x^2 - \sqrt{5}x + 1 = 0\). This can be compared to the standard form \(ax^2 + bx + c = 0\) where \(a = 1\), \(b = -\sqrt{5}\), and \(c = 1\).
02

Use the quadratic formula

To find the solutions for \(x\), use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Plugging in the values from the equation, \(a = 1\), \(b = -\sqrt{5}\), and \(c = 1\).
03

Calculate the discriminant

We begin by calculating the discriminant \(b^2 - 4ac\). Here it is: \((-\sqrt{5})^2 - 4(1)(1) = 5 - 4 = 1\). Since the discriminant is positive, there are two real solutions.
04

Solve for the roots

Substitute the discriminant and values of \(a\) and \(b\) into the quadratic formula: \(x = \frac{-(-\sqrt{5}) \pm \sqrt{1}}{2 \cdot 1}\). Simplifying, \(x = \frac{\sqrt{5} \pm 1}{2}\).
05

Express the solutions

The two solutions are \(x_1 = \frac{\sqrt{5} + 1}{2}\) and \(x_2 = \frac{\sqrt{5} - 1}{2}\). These are the real solutions of the quadratic equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Formula
To solve quadratic equations like the one given, \(x^2 - \sqrt{5}x + 1 = 0\), we often use a powerful tool known as the quadratic formula. The quadratic formula is applicable to any quadratic equation in the standard form \(ax^2 + bx + c = 0\). By substituting a, b, and c from our equation into the formula:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
You can determine the values of \(x\), which are the solutions of the equation. In this formula:
  • \(a\) is the coefficient of \(x^2\)
  • \(b\) is the coefficient of \(x\)
  • \(c\) is the constant term
For our particular equation, we have \(a = 1\), \(b = -\sqrt{5}\), and \(c = 1\). Substituting these into the quadratic formula allows us to move to discriminant calculation.
Real Solutions
Determining whether a quadratic equation has real solutions depends on the value of the discriminant. The discriminant is a part of the quadratic formula under the square root, \(b^2 - 4ac\). The value of the discriminant tells us:
  • If \(b^2 - 4ac > 0\), there are two distinct real solutions.
  • If \(b^2 - 4ac = 0\), there is exactly one real solution, sometimes called a double root.
  • If \(b^2 - 4ac < 0\), the equation does not have real solutions but rather two complex solutions.
In the equation \(x^2 - \sqrt{5}x + 1 = 0\), the discriminant was calculated to be 1 which is greater than 0. This result confirms that the equation has two distinct real solutions.
Discriminant Calculation
Calculating the discriminant is a crucial step in the process of solving quadratic equations. The discriminant formula is given by \(b^2 - 4ac\). For our equation, rewinding to the values \(a = 1\), \(b = -\sqrt{5}\), and \(c = 1\), let's calculate this step by step:

Step-by-step Calculation

- Firstly, compute \((-\sqrt{5})^2\) which results in 5.- Then, calculate \(4 \, \cdot \, 1 \, \cdot \, 1\), which equals 4.- Subtracting these gives you the discriminant: \(5 - 4 = 1\).This positive discriminant value tells us that the quadratic equation will have two real solutions. This process ensures a clear understanding of the nature of the roots without even solving for them directly.

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Most popular questions from this chapter

Use the Difference of Squares Formula to factor \(17^{2}-16^{2}\). Notice that it is easy to calculate the factored form in your head, but not so easy to calculate the original form in this way. Evaluate each expression in your head: (a) \(528^{2}-527^{2}\) (b) \(122^{2}-120^{2}\) (c) \(1020^{2}-1010^{2}\) Now use the Special Product Formula $$(A+B)(A-B)=A^{2}-B^{2}$$ to evaluate these products in your head: (d) \(79 \cdot 51\) (e) \(998 \cdot 1002\)

Factor the trinomial. $$2(a+b)^{2}+5(a+b)-3$$

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