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Chapter 11: Problem 5

Solve the quadratic equation \(5 x^{2}-11 x+13=0\)

Short Answer

Expert verified
Answer: The solutions to the quadratic equation are: \(x_1 = \frac{11 + \sqrt{-139}}{10}\) and \(x_2 = \frac{11 - \sqrt{-139}}{10}\).

Step by step solution

01

Identify coefficients a, b, and c

In the given quadratic equation \(5x^{2} - 11x + 13 = 0\), the coefficients are: a = 5 b = -11 c = 13
02

Substitute the coefficients into the quadratic formula

Now, we can substitute the values of a, b, and c into the quadratic formula: x = \(\frac{-(-11) \pm \sqrt{(-11)^2 - 4(5)(13)}}{2(5)}\)
03

Simplify the expression

Let's simplify the expression: x = \(\frac{11 \pm \sqrt{121 - 260}}{10}\) x = \(\frac{11 \pm \sqrt{-139}}{10}\)
04

Write the final solution

Since there are no real solutions due to having the negative number inside the square root, the final solution remains in a complex form: x = \(\frac{11 \pm \sqrt{-139}}{10}\) The solutions to the quadratic equation \(5x^{2} - 11x + 13 = 0\) are: \(x_1 = \frac{11 + \sqrt{-139}}{10}\) and \(x_2 = \frac{11 - \sqrt{-139}}{10}\).

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Key Concepts

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

Quadratic Formula
The quadratic formula is a powerful tool for finding the roots of a quadratic equation, which is any equation that can be written in the form \(ax^2 + bx + c = 0\). It's especially useful when the equation does not factor neatly. The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this formula, \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation. Here's a short breakdown of its use:
  • Start by identifying \(a\), \(b\), and \(c\) from your equation.
  • Substitute these values into the quadratic formula.
  • Solve for \(x\) by performing the necessary arithmetic operations.
Using the quadratic formula can help uncover both real and complex solutions, making it a very versatile method.
When applying it, ensure careful arithmetic to avoid any errors, especially when dealing with negative numbers or complex numbers.
Complex Numbers
Complex numbers arise when dealing with the square root of a negative number. They allow us to extend our number system beyond the real numbers. A complex number is generally represented as \(a + bi\), where:
  • \(a\) is the real part.
  • \(b\) is the imaginary part.
  • \(i\) is the imaginary unit, with the property \(i^2 = -1\).
In quadratic equations, if the term under the square root (the discriminant) is negative, this leads to complex solutions. For example, in the equation \(5x^2 - 11x + 13 = 0\), we ended up with \(\sqrt{-139}\), interpreted as a complex number.
This means the equation has solutions that are not real, but rather of the form \(a + bi\). Learning to work with complex numbers broadens your ability to solve various equations that you might encounter.
Solutions of Equations
Finding the solutions of an equation means identifying values of \(x\) that satisfy the equation. For quadratic equations, these solutions may be real numbers or complex numbers.
The quadratic formula provides these solutions directly.To understand solutions:
  • If the discriminant \((b^2 - 4ac)\) is positive, the equation has two distinct real solutions.
  • If it is zero, both solutions are real and identical (a perfect square).
  • If it is negative, the solutions are complex.
In our example, because the discriminant is \(-139\), the solutions are complex:\
\(x_1 = \frac{11 + \sqrt{-139}}{10}\) and \(x_2 = \frac{11 - \sqrt{-139}}{10}\). This tells us that the values of \(x\) that make \(5x^2 - 11x + 13 = 0\) true do not exist on the real number line but rather have an imaginary component.

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