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Chapter 2: Problem 9

For the following exercises, solve the quadratic equation by factoring. $$ 6 x^{2}+17 x+5=0 $$

Short Answer

Expert verified
The solutions are \(x = -\frac{1}{3}\) and \(x = -\frac{5}{2}\).

Step by step solution

01

Identify the Quadratic Equation

The given quadratic equation is \(6x^2 + 17x + 5 = 0\). It is in the standard form \(ax^2 + bx + c = 0\), where \(a = 6\), \(b = 17\), and \(c = 5\).
02

Find Two Numbers that Multiply to ac and Add to b

Calculate \(ac\), which is \(a \times c = 6 \times 5 = 30\). Find two numbers that multiply to 30 and add to 17. These numbers are 15 and 2, since \(15 \times 2 = 30\) and \(15 + 2 = 17\).
03

Rewrite the Middle Term

Rewrite the quadratic equation using the numbers found in Step 2: \(6x^2 + 15x + 2x + 5 = 0\). This replaces the 17x with 15x and 2x.
04

Factor by Grouping

Group the terms to factor by grouping: \((6x^2 + 15x) + (2x + 5) = 0\).Factor out the greatest common factor from each group: - From \(6x^2 + 15x\), factor out \(3x\) to get \(3x(2x + 5)\).- From \(2x + 5\), factor out 1 to get \(1(2x + 5)\).The equation becomes \(3x(2x + 5) + 1(2x + 5) = 0\).
05

Factor Out the Common Binomial

Recognize the common binomial factor \((2x + 5)\) in both terms: \((3x + 1)(2x + 5) = 0\).
06

Solve for x

Set each factor equal to zero and solve for \(x\): 1. \(3x + 1 = 0\) leads to \(3x = -1\), thus \(x = -\frac{1}{3}\). 2. \(2x + 5 = 0\) leads to \(2x = -5\), thus \(x = -\frac{5}{2}\).

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Key Concepts

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

Factoring
Factoring is a method used to solve quadratic equations by expressing them as a product of simpler polynomials. It's a technique where you find two numbers that satisfy a specific set of conditions. For a quadratic equation of the form \(ax^2 + bx + c = 0\), you look for two numbers that multiply to \(ac\) (the product of the coefficient of \(x^2\) and the constant term) and add up to \(b\) (the coefficient of \(x\)).
  • Example: In the quadratic equation \(6x^2 + 17x + 5 = 0\), notice \(a = 6\), \(b = 17\), and \(c = 5\). The product \(ac = 30\). A pair such as 15 and 2 can multiply to 30 and add to 17.
  • You then rewrite the middle term, \(17x\), using these numbers to split it into two terms: \(6x^2 + 15x + 2x + 5\).
  • Group the first two terms and the last two terms, i.e., \((6x^2 + 15x) + (2x + 5) = 0\), allowing you to factor by grouping.
  • Finally, factor each group and then factor out the common binomial, leading you to find \(x\) values that satisfy the original quadratic equation.
This method is efficient for quadratics that can be easily decomposed into factorable parts. However, not all quadratic equations are factorable using integers, which is when other methods, like the quadratic formula, can be useful.
Quadratic Formula
The quadratic formula is a universal method used to solve any quadratic equation, regardless of its ability to be factored easily. The formula is derived from completing the square on the general form of a quadratic equation. If your quadratic equation is in the form \(ax^2 + bx + c = 0\), the solutions can be found using:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
  • The term under the square root, \(b^2 - 4ac\), is called the discriminant. It determines the nature of the roots:
    • If it is positive, there are two distinct real roots.
    • If it is zero, there's exactly one real root or a repeated real root.
    • If it is negative, the quadratic equation has no real roots, but has two complex numbers as roots.
  • Using our example \(6x^2 + 17x + 5 = 0\), if factoring seems tricky, you can readily apply the quadratic formula to find \(x\).
It is important because it works for every quadratic equation, providing a straightforward solution even when other techniques are impractical.
Standard Form of Quadratic Equation
Quadratic equations are expressed in the standard form \(ax^2 + bx + c = 0\). This structure is essential for identifying the coefficients \(a\), \(b\), and \(c\), which play a key role in solving the equation.
  • 'A' is the coefficient of \(x^2\), 'B' is the coefficient of \(x\), and 'C' is the constant term.
  • In our exercise, the equation \(6x^2 + 17x + 5 = 0\) is in this standard form with \(a = 6\), \(b = 17\), and \(c = 5\).
  • Knowing these coefficients is crucial, whether you're solving the equation by factoring, using the quadratic formula, or other methods.
This standardized structure allows students to easily apply various mathematical techniques and check the nature of the equation's solutions.

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Most popular questions from this chapter

For the following exercises, solve for the unknown variable. $$ x^{-2}-x^{-1}-12=0 $$

For the following exercises, use the model for the period of a pendulum, \(T\), such that \(T=2 \pi \sqrt{\frac{L}{g}}\), where the length of the pendulum is \(L\) and the acceleration due to gravity is g. If the acceleration due to gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) and the period equals \(1 \mathrm{~s}\), find the length to the nearest \(\mathrm{cm}(100 \mathrm{~cm}=1 \mathrm{~m})\).

Solve \(x^{2}-x>12\)

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