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Chapter 5: Problem 2

Find two numbers whose product is (a) and whose sum is (b). a. 6 b. -5

Short Answer

Expert verified
The two numbers that are required are -2 and -3

Step by step solution

01

Formulate The Quadratic Equation

Given that the sum of two numbers is -5 and the product is 6, the two numbers are the roots of a quadratic equation. We express this as \( x^2 - Bx + C = 0 \) where \( B \) is the sum of the roots and \( C \) is the product of the roots. Substitute the given values into the equation: \( x^2 - (-5)x + 6 = 0 \) Simplify the equation to get \( x^2 + 5x + 6 = 0 \)
02

Solve The Quadratic Equation

The quadratic equation \( x^2 + 5x + 6 = 0 \) can be factored into \( (x + 2)(x + 3) = 0 \). By setting each factor equal to zero, we can find the roots of the equation.
03

Find The Roots

The first factor, \( x + 2 = 0 \), gives us the first root \( x = -2 \). The second factor, \( x + 3 = 0 \), gives us the second root \( x = -3 \).

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Key Concepts

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

Quadratic Formula
In math, a quadratic equation is expressed in the form \( ax^2 + bx + c = 0 \). The roots of this equation can be found using the quadratic formula: \[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]This powerful tool allows us to find solutions or roots for any quadratic equation. Here, \( a \), \( b \), and \( c \) are coefficients that you get from the equation. The symbol \( \pm \) means you can have two solutions.
Keep in mind:
  • \( b^2 - 4ac \) is called the discriminant.
  • If the discriminant is positive, you get two real roots.
  • If it's zero, there's exactly one root.
  • If negative, roots are complex or imaginary.
While the quadratic formula is very useful, sometimes it's quicker to solve a quadratic equation by factoring if it's easily factorable, as seen next.
Factoring Polynomials
Factoring is like taking a number apart to see what numbers multiply together to come up with the original. In this case, we're interested in factoring polynomials. Specifically, our quadratic polynomial is \( x^2 + 5x + 6 \).
The goal in factoring is to express the polynomial as a product of simpler expressions, called factors.
For \( x^2 + 5x + 6 \), look for two numbers that multiply to 6 (the constant term) and add up to 5 (the coefficient of \( x \)). These numbers are 2 and 3.
Thus, you write the equation as:
  • \((x + 2)(x + 3) = 0\)
This tells us the expression is fully factored. This makes it easier to find the solutions.
Solving Equations
Once you have factored a polynomial, solving it becomes straightforward. For the equation \((x + 2)(x + 3) = 0\), use the zero product property. This property states if a product of multiple factors equals zero, then at least one of the factors must be zero.
So, set each factor equal to zero:
  • \(x + 2 = 0\)
  • \(x + 3 = 0\)
Solving these equations individually gives the roots of the quadratic equation:
  • For \(x + 2 = 0\), subtract 2: \(x = -2\)
  • For \(x + 3 = 0\), subtract 3: \(x = -3\)
Therefore, the solutions, or roots, of the quadratic equation are \(x = -2\) and \(x = -3\). These solutions meet the problem's requirements鈥攖hey multiply to give 6 and add up to -5. This simple approach makes solving quadratic equations by factoring both efficient and effective.

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Most popular questions from this chapter

Factor. If the polynomial is prime, so indicate. $$5 x^{2}+2+x$$

Factor. $$a^{2}+6 a-16$$

Factor. If the polynomial is prime, so indicate. $$3 b^{2}+3 a^{2}-a b$$

Factor. Write each trinomial in descending powers of one variable, if necessary. If a polynomial is prime, so indicate. $$x^{2}-9 x-27$$

Factor. Write each trinomial in descending powers of one variable, if necessary. If a polynomial is prime, so indicate. $$a^{2}+2 a b+b^{2}-4$$
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