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Chapter 1: Problem 28

Solve. Some of your answers may involve \(i\). \(x^{2}+4 x=3\)

Short Answer

Expert verified
The solutions are \(-2 + \sqrt{7}\) and \(-2 - \sqrt{7}\).

Step by step solution

01

- Move all terms to one side

Start by moving all terms to one side of the equation to set it equal to zero: \ \[ x^{2} + 4x - 3 = 0 \]
02

- Identify coefficients

Identify the coefficients in the quadratic equation of the form \(ax^2 + bx + c = 0\): \ \[ a = 1, \ b = 4, \ c = -3 \]
03

- Use the quadratic formula

Use the quadratic formula to solve for \(x\): \ \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \ Substitute the coefficients: \ \[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-3)}}{2(1)} \]
04

- Simplify under the square root

Simplify the expression under the square root: \ \[ x = \frac{-4 \pm \sqrt{16 + 12}}{2} \ = \frac{-4 \pm \sqrt{28}}{2} \] \Since \sqrt{28} = 2\sqrt{7}, this further simplifies to: \ \[ x = \frac{-4 \pm 2\sqrt{7}}{2} \]
05

- Divide by 2

Divide each term in the numerator by 2: \ \[ x = -2 \pm \sqrt{7} \]

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Key Concepts

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

solving quadratic equations
A quadratic equation is any equation that can be written in the form: det with 3 terms: the square term, the linear term, and the constant. Here's what each term looks like in an equation: \[ ax^2 + bx + c = 0 \] where:
  • \(a\): the coefficient of the square term \(x^2\)
  • \(b\): the coefficient of the linear term \(x\)
  • \(c\): the constant.
To solve a quadratic equation, you want to find the value(s) of \(x\) that make the equation true. You generally start by moving all terms to one side of the equation so that one side equals zero. For instance, with the given problem: \[ x^2 + 4x = 3 \] you first rearrange it to: \[ x^2 + 4x - 3 = 0 \].
quadratic formula
One of the most powerful tools for solving quadratic equations is the quadratic formula. The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula allows you to solve any quadratic equation, provided you identify the coefficients \(a\), \(b\), and \(c\) from the equation. Let鈥檚 use our example: \[ x^2 + 4x - 3 = 0 \] Here, \(a = 1\), \(b = 4\), and \(c = -3\). Plugging these values into the formula gives: \[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-3)}}{2(1)} \] Simplifying inside the square root: \[ x = \frac{-4 \pm \sqrt{16 + 12}}{2} = \frac{-4 \pm \sqrt{28}}{2} \].
complex numbers
When solving quadratic equations, you might encounter a situation where the term under the square root becomes negative. This is where complex numbers come into play. The imaginary unit \(i\) is defined as \(i = \sqrt{-1}\). If we ever have to take the square root of a negative number, it helps to rewrite it using \(i\). For example, if you encounter \(\sqrt{-4}\), you can simplify it to \(2i\).In our given problem, we did not encounter a negative under the square root, but this concept is important to understand for more challenging problems.
simplifying square roots
Simplifying the square root is often a crucial step in solving quadratic equations. Let鈥檚 go back to the expression: \[ \frac{-4 \pm \sqrt{28}}{2} \] First, note that \(\sqrt{28}\) can be broken down further: \[ \sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7} \] So the equation changes to: \[ \frac{-4 \pm 2\sqrt{7}}{2} \] Next, divide each term in the numerator by 2: \[ x = -2 \pm \sqrt{7} \] Now, the solutions to the quadratic equation are: \[ x = -2 + \sqrt{7} \quad \text{and} \quad x = -2 - \sqrt{7} \].

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Use a calculator to find the radian measure of an acute angle whose trigonometric function is given. \(\tan t=1.26\)

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