Chapter 5: Problem 79
Solve each equation. $$ 0=d^{2}-5 d-66 $$
Short Answer
Expert verified
The solutions are \( d = -6 \) and \( d = 11 \).
Step by step solution
01
Understand the Equation Structure
The given equation is a quadratic equation in the standard form: \( ax^2 + bx + c = 0 \). In our equation, \( a = 1 \), \( b = -5 \), and \( c = -66 \). Our task is to find the values of \( d \) that satisfy the equation \( d^2 - 5d - 66 = 0 \).
02
Apply Factoring
We will attempt to factor the quadratic equation. We need to identify two numbers that multiply to \( ac \) (which is \( 1 \times (-66) = -66 \)) and add to \( b \) (which is \(-5\)). The numbers \(-11\) and \(6\) work because \(-11 \times 6 = -66\) and \(-11 + 6 = -5\).
03
Rewrite and Factor the Equation
Using the numbers identified, rewrite the middle term of the equation. The equation becomes: \[ d^2 - 11d + 6d - 66 = 0 \]. Now factor by grouping: \[ (d^2 - 11d) + (6d - 66) = 0 \]. Factor out the common factors: \[ d(d - 11) + 6(d - 11) = 0 \].
04
Set Factors to Zero
Notice \((d - 11)\) is a common factor. Factor it out: \[ (d + 6)(d - 11) = 0 \]. Set each factor equal to zero: \( d + 6 = 0 \) and \( d - 11 = 0 \).
05
Solve for \(d\)
Solve each linear equation to find \(d\): For \( d + 6 = 0 \), subtract 6: \( d = -6 \).For \( d - 11 = 0 \), add 11: \( d = 11 \). Thus the solutions are \( d = -6 \) and \( d = 11 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring
Factoring is a crucial method when solving quadratic equations. It involves expressing the equation in a product of simpler expressions. For the given quadratic equation \[ 0 = d^2 - 5d - 66, \]our goal is to find two numbers that multiply to the product of the coefficient of \(d^2\) (which is 1) and the constant term (which is -66). These two numbers must also add to the coefficient of the middle term, \(d\), which is -5.
- First, calculate the product: \(1 imes -66 = -66\).
- Next, find two numbers that multiply to -66 and add to -5. The numbers -11 and 6 meet these conditions \( (-11 imes 6 = -66\text{ and } -11 + 6 = -5)\).
Quadratic Formula
In cases where factoring is difficult, the quadratic formula offers a reliable alternative. For a quadratic equation in the form \[ ax^2 + bx + c = 0, \]the solutions \(x\) can be found using:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]This formula derives from completing the square technique and is useful in finding solutions when quadratic equations have complex or irrational roots.
- Calculate the discriminant: \(b^2 - 4ac\). It indicates the nature of the roots:
- If positive, there are two distinct real roots.
- If zero, there's exactly one real root.
- If negative, the roots are complex.
- Substitute the values for \(a, b,\) and \(c\) into the formula. For our exercise, \(a = 1, b = -5, c = -66\).
Polynomial Equations
Quadratic equations are a specific type of polynomial equations, where the highest exponent of the variable is two. More generally, a polynomial equation can include variables raised to any integer power. Quadratics themselves fall into the category of second-degree polynomials:\[ ax^2 + bx + c = 0. \]The challenge is usually finding the roots, i.e., the values of the variable that make the equation true.
- The degree of the polynomial determines the possible number of solutions. A quadratic equation can have up to two distinct roots.
- These can often be determined by factoring, using the quadratic formula, or graphically finding the intercepts with the x-axis.
- In the context of our exercise, identifying the quadratic form was the first step in solving the equation for real-world applications.