Chapter 11: Problem 22
Use the quadratic formula to solve each equation. These equations have real number solutions only. See Examples I through 3. $$ 7 p(p-2)+2(p+4)=3 $$
Short Answer
Expert verified
The solutions are \( p = 1 \) and \( p = \frac{5}{7} \).
Step by step solution
01
Expand and Simplify the Equation
The given equation is \( 7p(p-2) + 2(p+4) = 3 \). First, expand both terms: - \( 7p(p-2) = 7p^2 - 14p \) and - \( 2(p+4) = 2p + 8 \).Combine these terms: \( 7p^2 - 14p + 2p + 8 = 3 \).Simplify: \( 7p^2 - 12p + 8 = 3 \).
02
Set the Equation to Zero
Subtract 3 from both sides to get the equation in standard form: \[ 7p^2 - 12p + 8 - 3 = 0 \]. Simplify this to \[ 7p^2 - 12p + 5 = 0 \].
03
Identify Coefficients
Identify the coefficients in the quadratic equation \( 7p^2 - 12p + 5 = 0 \).- \( a = 7 \)- \( b = -12 \)- \( c = 5 \).
04
Apply the Quadratic Formula
The quadratic formula is \[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].Substitute the coefficients into the formula: \[ p = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 7 \cdot 5}}{2 \cdot 7} \].
05
Calculate the Discriminant
Calculate the discriminant \( b^2 - 4ac = (-12)^2 - 4 \cdot 7 \cdot 5 \).Compute: \( 144 - 140 = 4 \).
06
Solve for p
Since the discriminant is 4, compute the roots:\[ p = \frac{12 \pm \sqrt{4}}{14} \].Simplify \( \sqrt{4} = 2 \): \( p = \frac{12 + 2}{14} \) and \( p = \frac{12 - 2}{14} \).Calculate: \( p = \frac{14}{14} = 1 \) \( p = \frac{10}{14} = \frac{5}{7} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Discriminant
The discriminant is a key concept when dealing with quadratic equations. It's represented by the expression \( b^2 - 4ac \), which you find inside the square root of the quadratic formula. The discriminant helps determine the nature of the solutions of a quadratic equation.
- If the discriminant is positive, the quadratic equation has two distinct real number solutions.
- If the discriminant is zero, there is exactly one real number solution, which is also called a repeated root.
- If the discriminant is negative, the quadratic equation does not have real number solutions; instead, it has two complex solutions.
Quadratic Equation
A quadratic equation is a type of polynomial equation of degree two, generally expressed in the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are coefficients, and \( a \) is not zero. Such equations can graph as a parabola in a coordinate plane. The axis of symmetry and the vertex are important features of this parabola.
- "\( a \)" is the coefficient of the term \( x^2 \) and it determines the direction of the parabola. If \( a > 0 \), the parabola opens upwards. If \( a < 0 \), it opens downwards.
- "\( b \)" is the coefficient of the term \( x \) and affects the parabola's symmetry.
- "\( c \)" is the constant term and it places the vertex on the y-axis.
Real Number Solutions
Real number solutions arise when the solutions of an equation are real numbers rather than complex numbers. When we use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), it is crucial to look at the discriminant to predict the nature of the solutions before diving into solving it.
Let's analyze the quadratic formula for real solutions:
Let's analyze the quadratic formula for real solutions:
- The formula will result in real solutions if the expression inside the square root, \( b^2 - 4ac \), is zero or positive.
- If \( \sqrt{b^2 - 4ac} \) equals \( 0 \), you'll get one real solution because the equation simplifies to just one value.
- If \( \sqrt{b^2 - 4ac} \) is a positive number, it contributes to "+/-," leading to two distinct real solutions.