Chapter 7: Problem 9
Solve the given quadratic equations using the quadratic formula. If there are no real roots, state this as the answer. Exercises \(3-6\) are the same as Exercises \(13-16\) of Section 7.2. $$v^{2}=15-2 v$$
Short Answer
Expert verified
The solutions are \( v = 3 \) and \( v = -5 \).
Step by step solution
01
Write the equation in standard form
A quadratic equation should be in the standard form of \( ax^2 + bx + c = 0 \). Starting with \( v^2 = 15 - 2v \), move all terms to one side to get: \( v^2 + 2v - 15 = 0 \). Now, identify \( a = 1 \), \( b = 2 \), and \( c = -15 \).
02
Apply the Quadratic Formula
The quadratic formula is given by \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 1 \), \( b = 2 \), and \( c = -15 \) into this formula: \( v = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1} \).
03
Calculate the discriminant
The discriminant \( b^2 - 4ac \) determines the nature of the roots. Compute it: \( 2^2 - 4 \cdot 1 \cdot (-15) = 4 + 60 = 64 \). Since the discriminant is positive, there are two distinct real roots.
04
Simplify under the square root
Calculate the square root: \( \sqrt{64} = 8 \). This will be used in the quadratic formula.
05
Find the roots
Substitute \( \sqrt{64} = 8 \) into the quadratic formula: \( v = \frac{-2 \pm 8}{2} \). Calculate the two possible values: \( v = \frac{-2 + 8}{2} = 3 \) and \( v = \frac{-2 - 8}{2} = -5 \).
06
Check the final solution
Verify the solutions by substituting back into the original equation. For \( v = 3 \), \( v^2 = 9 \) and \( 15 - 2v = 9 \), which satisfy the equation. For \( v = -5 \), \( v^2 = 25 \) and \( 15 - 2v = 25 \), confirming both solutions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Quadratic Formula
Quadratic equations often take the form of a polynomial equation written as \( ax^2 + bx + c = 0 \).
The coefficients \( a \), \( b \), and \( c \) represent known values, where \( a eq 0 \).
Among different methods to solve quadratic equations, using the quadratic formula is a universal one.
The quadratic formula is expressed as:
By substituting the values of \( a \), \( b \), and \( c \) into the formula, students will find the solutions or roots of the equation.
If you are solving a quadratic equation like \( v^2 + 2v - 15 = 0 \), plugging the values directly into the formula will reveal your equation's roots.
The coefficients \( a \), \( b \), and \( c \) represent known values, where \( a eq 0 \).
Among different methods to solve quadratic equations, using the quadratic formula is a universal one.
The quadratic formula is expressed as:
- \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
By substituting the values of \( a \), \( b \), and \( c \) into the formula, students will find the solutions or roots of the equation.
If you are solving a quadratic equation like \( v^2 + 2v - 15 = 0 \), plugging the values directly into the formula will reveal your equation's roots.
Real Roots: What They Are and How to Find Them
Real roots are the solutions of the quadratic equation that are real numbers, not involving any imaginary part.
Depending on the equation, a quadratic can have two real roots, one real root, or none.
To determine if a quadratic equation has real roots, we use the discriminant \( b^2 - 4ac \):
Depending on the equation, a quadratic can have two real roots, one real root, or none.
To determine if a quadratic equation has real roots, we use the discriminant \( b^2 - 4ac \):
- If the discriminant is positive, as in this example where the value is \( 64 \),\( b^2 - 4ac > 0 \), the equation has two distinct real roots.
- If it is zero, \( b^2 - 4ac = 0 \), there is exactly one real root, often described as a repeated root.
- Where the discriminant is negative, \( b^2 - 4ac < 0 \), the equation has no real roots, only complex or imaginary roots.
The Discriminant: Gateway to Understanding Roots
In the quadratic formula, the discriminant \( b^2 - 4ac \) is the expression under the square root sign.
It plays a critical role in determining the number and type of roots for any quadratic equation.
Let's delve a little deeper:
When we calculate the discriminant, we're literally assessing the content of the term under the square root to see how it contributes to the overall solution.
It plays a critical role in determining the number and type of roots for any quadratic equation.
Let's delve a little deeper:
When we calculate the discriminant, we're literally assessing the content of the term under the square root to see how it contributes to the overall solution.
- If the discriminant equals zero, the roots are real and equal, signifying that the graph of the equation touches the x-axis at exactly one point.
- A positive discriminant, as in this problem where it resulted in \( 64 \), suggests two different x-intercepts representing two separate real numbers.