Chapter 3: Problem 26
USING EQUATIONS How many solutions does the system have? Explain your reasoning. $$ \begin{aligned} &y=7 x^2-11 x+9 \\ &y=-7 x^2+5 x-3 \end{aligned} $$ (A) 0 (B) 1 (C) 2 (D) 4
Short Answer
Expert verified
(C) 2
Step by step solution
01
Equate the two functions
Start by setting the two equations equal to each other to find the values of \(x\) where the equations intersect: \(7x^2 - 11x + 9 = -7x^2 + 5x - 3\)
02
Simplify the equation
Combine the similar terms in the left and right side of the equation, which gives \(14x^2 -16x +12 = 0\)
03
Solve for the roots
The roots of the equation can be found by either factoring, completing the square or using the quadratic formula. Here, the quadratic formula \(x = [-b ± sqrt(b^2 - 4ac)] / (2a)\) will be used. Where \(a = 14, b = -16, c = 12\). By substituting these values in, we obtain \(x = 3/7 , 4/7\). Therefore, there are two solutions for the system of equations
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations are a type of polynomial equation, characterized by having an exponent of 2 for at least one of its terms. The general form is written as \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are real numbers, and \( a eq 0 \). The quadratic equation is fundamental in algebra because it allows us to model various real-world scenarios involving area, projectile motion, and more. In practical applications, solving quadratic equations means finding the value(s) of \( x \) that make the equation true. These values are known as the roots of the equation.
Quadratic equations can have:
Quadratic equations can have:
- Two distinct solutions
- One repeated solution
- No real solutions
Intersecting Solutions
When dealing with systems of equations, finding intersecting solutions means identifying the points at which the graphs of the equations meet. For a system involving quadratic equations, these intersections illustrate when and where two parabolic curves cross each other on a graph.
In the context of the provided exercise, the system of equations is given by two quadratic functions. To determine their intersecting solutions, you set the equations equal to each other: \( 7x^2 - 11x + 9 = -7x^2 + 5x - 3 \). This setup helps us find the x-values where both equations yield the same y-value.
Solving this results in a new quadratic equation that gives the x-coordinates of the intersection points. By substituting back into either original equation, the corresponding y-values are found, marking the precise intersection points on the graph.
In the context of the provided exercise, the system of equations is given by two quadratic functions. To determine their intersecting solutions, you set the equations equal to each other: \( 7x^2 - 11x + 9 = -7x^2 + 5x - 3 \). This setup helps us find the x-values where both equations yield the same y-value.
Solving this results in a new quadratic equation that gives the x-coordinates of the intersection points. By substituting back into either original equation, the corresponding y-values are found, marking the precise intersection points on the graph.
Quadratic Formula
The quadratic formula is a universal tool to solve any quadratic equation of the form \( ax^2 + bx + c = 0 \). It provides a way to find the roots, even when they cannot be easily factored. The formula is written as:
In our exercise, applying the quadratic formula to the equation \( 14x^2 -16x +12 = 0 \) after simplification results in two solutions:\( x = \frac{3}{7} \) and \( x = \frac{4}{7} \). These solutions represent the x-values where the original quadratic equations intersect.
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In our exercise, applying the quadratic formula to the equation \( 14x^2 -16x +12 = 0 \) after simplification results in two solutions:\( x = \frac{3}{7} \) and \( x = \frac{4}{7} \). These solutions represent the x-values where the original quadratic equations intersect.
Algebraic Simplification
Algebraic simplification is the process of making an equation more manageable by combining like terms and reducing the equation into a simpler form without changing its solutions. This is crucial when solving systems of equations, especially those involving quadratics.
In our exercise, we start by setting the two quadratic equations equal to each other.\( 7x^2 - 11x + 9 = -7x^2 + 5x - 3 \). By moving terms around and simplifying, we combine like terms, yielding the simplified equation \( 14x^2 -16x +12 = 0 \). Through simplification, complex problems become easier to handle and solve, enabling us to apply methods like the quadratic formula more effectively. This step is particularly essential because it reduces potential errors and helps clearly identify the roots of the equation.
In our exercise, we start by setting the two quadratic equations equal to each other.\( 7x^2 - 11x + 9 = -7x^2 + 5x - 3 \). By moving terms around and simplifying, we combine like terms, yielding the simplified equation \( 14x^2 -16x +12 = 0 \). Through simplification, complex problems become easier to handle and solve, enabling us to apply methods like the quadratic formula more effectively. This step is particularly essential because it reduces potential errors and helps clearly identify the roots of the equation.