/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 130 Solve the equation: x^{2}+2 \sqr... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 130

Solve the equation: x^{2}+2 \sqrt{3} x-9=0

Short Answer

Expert verified
The solutions to this quadratic equation are \( x = \sqrt{3} \) and \( x = -3\sqrt{3} \).

Step by step solution

01

Identifying Coefficients

In the equation \( x^{2}+2 \sqrt{3} x-9=0 \), the coefficient 'a' is 1, 'b' is \(2 \sqrt{3}\), and 'c' is -9.
02

Calculating Under the Square Root in Quadratic Formula

Utilize the formula \(b^{2}-4ac\). Substituting values we get \((2 \sqrt{3})^{2}-4*1*(-9)\), which simplifies to \(12+36 = 48\).
03

Using Quadratic Formula

Now substitute these values in the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \) and calculate the value of 'x'. This would yield \( x = \frac{-2 \sqrt{3} \pm \sqrt{48}}{2} \). To simplify this, we calculate \( x1 = \frac{-2 \sqrt{3} + \sqrt{48}}{2} \), and \( x2 = \frac{-2 \sqrt{3} - \sqrt{48}}{2} \). Simplifying further gives the solutions \( x = \sqrt{3} \) and \( x = -3\sqrt{3} \).

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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 solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The formula provides the solution for \( x \) in terms of the coefficients of the quadratic equation:
  • \( a \) is the coefficient of \( x^2 \)
  • \( b \) is the coefficient of \( x \)
  • \( c \) is the constant term
The quadratic formula itself is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula gives two solutions for \( x \), corresponding to the \( + \) and \( - \) operations, known as the roots of the equation. Using this formula, you can find the exact solutions of any quadratic equation, assuming the calculation under the square root (the discriminant) is non-negative.
Solving Quadratic Equations
To solve a quadratic equation using the quadratic formula, follow these steps:
  • Identify the coefficients \( a \), \( b \), and \( c \) from the equation \( ax^2 + bx + c = 0 \).
  • Calculate the discriminant \( b^2 - 4ac \).
  • Substitute \( a \), \( b \), and the discriminant into the quadratic formula.
  • Compute the solutions for \( x \) using both the positive and negative values in the formula.
In the given exercise, the equation is \( x^2 + 2\sqrt{3}x - 9 = 0 \). Using the quadratic formula, we substitute \( a = 1 \), \( b = 2\sqrt{3} \), and \( c = -9 \). Solving this, you find the roots \( x = \sqrt{3} \) and \( x = -3\sqrt{3} \). These represent the points where the quadratic function crosses the x-axis.
Radicals in Algebra
Radicals are expressions that involve roots, such as square roots, cube roots, etc. In algebra, radicals are used to simplify expressions and solve equations. When you encounter a radical like \( \sqrt{48} \), it can often be simplified. To simplify \( \sqrt{48} \):
  • Recognize that \( 48 = 16 \times 3 \)
  • This allows you to simplify \( \sqrt{48} \) as \( \sqrt{16} \times \sqrt{3} \) which becomes \( 4\sqrt{3} \)
Simplifying radicals can make it easier to solve equations and understand the nature of the solutions. In the quadratic formula, simplifying the square root term is a crucial step in expressing the roots of the equation in their simplest form. This process ensures clarity when interpreting the solutions of quadratic equations, whether they have real or complex roots.

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Most popular questions from this chapter

In Exercises \(29-44,\) perform the indicated operations and write the result in standard form. $$\sqrt{-12}(\sqrt{-4}-\sqrt{2})$$

In Exercises \(1-16,\) solve and check each linear equation. $$ 7 x-5=72 $$

Which one of the following is true? a. The solution set of \(x^{2}>25\) is \((5, \infty)\) b. The inequality \(\frac{x-2}{x+3}<2\) can be solved by multiplying both sides by \(x+3\), resulting in the equivalent inequality \(x-2<2(x+3)\) c. \((x+3)(x-1) \geq 0\) and \(\frac{x+3}{x-1} \geq 0\) have the same solution set. d. None of these statements is true.

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x-4}{x+3}>0 $$

In Exercises \(9-20,\) find each product and write the result in standard form. $$(3+5 i)(3-5 i)$$
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