Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 11: Problem 75
Approximate all real zeros of each function to the nearest hundredth. \(f(x)=-\sqrt{15} x^{4}-\sqrt{3} x^{2}+7\)
Short Answer
Expert verified
The approximated real zeros are \( x \approx \pm 0.68 \).
Step by step solution
01
- Identify the function
The given function is \[ f(x) = -\sqrt{15} x^{4} - \sqrt{3} x^{2} + 7 \]We aim to find the values of x where this function equals zero.
02
- Set the function equal to zero
To find the zeros, set the function equal to zero:\[ -\sqrt{15} x^{4} - \sqrt{3} x^{2} + 7 = 0 \]
03
- Substitute \( y = x^2 \)
Let \( y = x^2 \). This substitution simplifies the equation to a quadratic form:\[ -\sqrt{15} y^{2} - \sqrt{3} y + 7 = 0 \]
04
- Use the quadratic formula
Solve the quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = -\sqrt{15} \), \( b = -\sqrt{3} \), and \( c = 7 \).First, compute the discriminant: \[ \Delta = b^2 - 4ac = ( -\sqrt{3})^2 - 4 ( -\sqrt{15}) ( 7 ) = 3 + 28 \sqrt{15} \]
05
- Find the roots for y
Calculate the roots using the formula:\[ y = \frac{ -\sqrt{3} \pm \sqrt{3 + 28 \sqrt{15}}}{2 ( -\sqrt{15})} \].The roots are approximately:\[ y_1 \approx -1.23 \text{ (not possible as } y = x^2\text{ must be non-negative)} \]\[ y_2 \approx 0.47 \]
06
- Calculate x values
Solve for x from the feasible \( y \) value:\[ x^2 = 0.47 \]Taking the square root, we get two potential solutions:\[ x = \pm \sqrt{0.47} \approx \pm 0.68 \]
07
- Approximate to the nearest hundredth
Thus, the real zeros of the function are approximately \( x \approx \pm 0.68 \) when rounded to the nearest hundredth.
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 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a handy tool for finding the roots of a quadratic equation, which takes the general form: \[ ax^2 + bx + c = 0 \]The formula itself is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Using it might look tricky due to the symbols, but it simply helps us find the x-values where the quadratic function equals zero. These x-values are called the 'roots' or 'zeros' of the function.
Here's a quick breakdown of the steps:
Here's a quick breakdown of the steps:
- Identify coefficients: - Find the values of a, b, and c from your equation.
- Plug into the formula: - Insert these coefficients into the quadratic formula.
- Calculate the discriminant: - The value inside the square root \( b^2 - 4ac \) decides how many real roots we get.
- Solve for x: - Finally, solve the equation to get the x-values.
Polynomial Equations
Polynomial equations are expressions involving variables and coefficients, set to equal zero. They have various degrees based on the highest power of the variable. For instance, a polynomial of degree four, such as \[ f(x) = -\sqrt{15} x^{4} - \sqrt{3} x^{2} + 7 \], takes a bit more work to solve.
Here's why polynomial equations are important:
Here's why polynomial equations are important:
- Roots and Zeros: They help find the values of x where the function evaluates to zero, crucial for graphing intersections and solving real-world problems.
- Function Behavior: Understanding their roots helps to predict the behavior of the polynomial function over different intervals.
- Special Cases: They often simplify into quadratic forms that are easier to solve, as seen in our given problem where we substituted \( y = x^2 \).
Real Zeros Calculation
Calculating the real zeros of a polynomial function means finding the x-values where the function equals zero. Here's a straightforward approach:
\( y = 0.47 \), meaning \( x^2 = 0.47 \). Solving this gives us the real zeros \( x = \pm 0.68 \). This step-by-step breakdown is essential for understanding the calculation process thoroughly.
- Set the equation to zero: Adjust the expression so that it equals zero.
- Transform to simpler form: Substitute variables if needed to simplify into a known form, like we did with \( y = x^2 \).
- Apply solution method: Use the quadratic formula or factor the polynomial if possible.
- Evaluate the zeros: Solve for the real values using arithmetic or formulaic calculations.
\( y = 0.47 \), meaning \( x^2 = 0.47 \). Solving this gives us the real zeros \( x = \pm 0.68 \). This step-by-step breakdown is essential for understanding the calculation process thoroughly.
Discriminant
The discriminant of a quadratic equation, found inside the square root of the quadratic formula, helps determine the number and nature of the roots. It is given by: \[ \text{Discriminant} = b^2 - 4ac \]Here's its significance:
- Positive Discriminant: If \( \Delta > 0 \), the equation has two distinct real roots.
- Zero Discriminant: If \( \Delta = 0 \), the equation has exactly one real root (also called a repeated or double root).
- Negative Discriminant: If \( \Delta < 0 \), the equation has no real roots, only complex ones.
