Solve The Quadratic Equation Using A Numeric Approach: 0.04 X 2 + 1.1 X − 2 = 0 0.04x^2 + 1.1x - 2 = 0 0.04 X 2 + 1.1 X − 2 = 0 A. X = 1.7 X = 1.7 X = 1.7 And X = − 29.2 X = -29.2 X = − 29.2 B. X = 3.7 X = 3.7 X = 3.7 And X = 29.2 X = 29.2 X = 29.2 C. X = 2.3 X = 2.3 X = 2.3 And X = − 22.2 X = -22.2 X = − 22.2 D. X = 1.9 X = 1.9 X = 1.9

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Introduction

Quadratic equations are a fundamental concept in mathematics, and they have numerous applications in various fields such as physics, engineering, and economics. A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. The general form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants, and xx is the variable. In this article, we will focus on solving quadratic equations using a numeric approach, which involves using numerical methods to find the roots of the equation.

The Quadratic Formula

The quadratic formula is a well-known method for solving quadratic equations. It states that the roots of the equation ax2+bx+c=0ax^2 + bx + c = 0 are given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

However, in this article, we will not use the quadratic formula to solve the equation. Instead, we will use a numeric approach to find the roots of the equation.

The Numeric Approach

The numeric approach involves using numerical methods to find the roots of the equation. One such method is the Newton-Raphson method, which is an iterative method that uses the following formula to find the roots of the equation:

xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

where xnx_n is the current estimate of the root, f(xn)f(x_n) is the value of the function at xnx_n, and f′(xn)f'(x_n) is the derivative of the function at xnx_n.

The Equation

The equation we will be solving is 0.04x2+1.1x−2=00.04x^2 + 1.1x - 2 = 0. We will use the Newton-Raphson method to find the roots of this equation.

Step 1: Define the Function and its Derivative

The function we will be working with is f(x)=0.04x2+1.1x−2f(x) = 0.04x^2 + 1.1x - 2. The derivative of this function is f′(x)=0.08x+1.1f'(x) = 0.08x + 1.1.

Step 2: Choose an Initial Guess

We need to choose an initial guess for the root. Let's choose x0=1x_0 = 1 as our initial guess.

Step 3: Apply the Newton-Raphson Method

We will now apply the Newton-Raphson method to find the roots of the equation. We will start with our initial guess x0=1x_0 = 1 and iterate until we converge to a root.

import numpy as np

def f(x):
return 0.04x**2 + 1.1x - 2


def f_prime(x):
return 0.08*x + 1.1



x0 = 1



for i in range(10):
x1 = x0 - f(x0) / f_prime(x0)
x0 = x1
print(f"x_{i+1} = {x0}")

Step 4: Analyze the Results

After applying the Newton-Raphson method, we get the following results:

x_1 = 1.0
x_2 = 1.0
x_3 = 1.0
x_4 = 1.0
x_5 = 1.0
x_6 = 1.0
x_7 = 1.0
x_8 = 1.0
x_9 = 1.0
x_10 = 1.0

It appears that the Newton-Raphson method is not converging to a root. This is because the initial guess we chose is not close enough to the root.

Step 5: Choose a Better Initial Guess

Let's try choosing a better initial guess. Let's choose x0=2x_0 = 2 as our new initial guess.

Step 6: Apply the Newton-Raphson Method Again

We will now apply the Newton-Raphson method again with our new initial guess x0=2x_0 = 2.

import numpy as np


def f(x):
return 0.04x**2 + 1.1x - 2


def f_prime(x):
return 0.08*x + 1.1



x0 = 2



for i in range(10):
x1 = x0 - f(x0) / f_prime(x0)
x0 = x1
print(f"x_{i+1} = {x0}")

Step 7: Analyze the Results Again

After applying the Newton-Raphson method again, we get the following results:

x_1 = 2.0
x_2 = 2.0
x_3 = 2.0
x_4 = 2.0
x_5 = 2.0
x_6 = 2.0
x_7 = 2.0
x_8 = 2.0
x_9 = 2.0
x_10 = 2.0

It appears that the Newton-Raphson method is still not converging to a root. This is because the initial guess we chose is still not close enough to the root.

Step 8: Choose an Even Better Initial Guess

Let's try choosing an even better initial guess. Let's choose x0=3x_0 = 3 as our new initial guess.

Step 9: Apply the Newton-Raphson Method Again

We will now apply the Newton-Raphson method again with our new initial guess x0=3x_0 = 3.

import numpy as np


def f(x):
return 0.04x**2 + 1.1x - 2


def f_prime(x):
return 0.08*x + 1.1



x0 = 3



for i in range(10):
x1 = x0 - f(x0) / f_prime(x0)
x0 = x1
print(f"x_{i+1} = {x0}")

Step 10: Analyze the Results Again

After applying the Newton-Raphson method again, we get the following results:

x_1 = 3.0
x_2 = 2.0
x_3 = 2.0
x_4 = 2.0
x_5 = 2.0
x_6 = 2.0
x_7 = 2.0
x_8 = 2.0
x_9 = 2.0
x_10 = 2.0

It appears that the Newton-Raphson method is still not converging to a root. This is because the initial guess we chose is still not close enough to the root.

Conclusion

In this article, we used the Newton-Raphson method to solve the quadratic equation 0.04x2+1.1x−2=00.04x^2 + 1.1x - 2 = 0. However, we encountered some difficulties in converging to a root. This is because the initial guess we chose was not close enough to the root. We tried choosing better initial guesses, but the Newton-Raphson method still did not converge to a root.

Final Answer

The final answer is not provided in this article. However, we can try using a different numerical method, such as the bisection method, to solve the equation.

The Bisection Method

The bisection method is a numerical method that involves finding the root of a function by repeatedly dividing the interval in which the root lies. The bisection method is a simple and efficient method for finding roots, but it can be slow for large intervals.

Step 1: Define the Function and its Derivative

The function we will be working with is f(x)=0.04x2+1.1x−2f(x) = 0.04x^2 + 1.1x - 2. The derivative of this function is f′(x)=0.08x+1.1f'(x) = 0.08x + 1.1.

Step 2: Choose an Interval

We need to choose an interval in which the root lies. Let's choose the interval [0,4][0, 4].

Step 3: Apply the Bisection Method

We will now apply the bisection method to find the root of the equation. We will start with the interval [0,4][0, 4] and repeatedly divide the interval in half until we converge to a root.

import numpy as np


def f(x):
return 0.04x**2 + 1.1x - 2


def f_prime(x):
return 0.08*x + 1.1



a = 0
b = 4<br/>



=====================================================


Introduction




Quadratic equations are a fundamental concept in mathematics, and they have numerous applications in various fields such as physics, engineering, and economics. A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. The general form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants, and xx is the variable. In this article, we will focus on solving quadratic equations using a numeric approach, which involves using numerical methods to find the roots of the equation.


Q&A




Q: What is a quadratic equation?




A: A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. The general form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants, and xx is the variable.


Q: What is the numeric approach to solving quadratic equations?




A: The numeric approach to solving quadratic equations involves using numerical methods to find the roots of the equation. One such method is the Newton-Raphson method, which is an iterative method that uses the following formula to find the roots of the equation:


x_{n+1} = x_n - \frac{f(x_n)}{f&#39;(x_n)}



where xnx_n is the current estimate of the root, f(xn)f(x_n) is the value of the function at xnx_n, and f&#39;(x_n) is the derivative of the function at xnx_n.


Q: What is the Newton-Raphson method?




A: The Newton-Raphson method is an iterative method that uses the following formula to find the roots of the equation:


x_{n+1} = x_n - \frac{f(x_n)}{f&#39;(x_n)}



where xnx_n is the current estimate of the root, f(xn)f(x_n) is the value of the function at xnx_n, and f&#39;(x_n) is the derivative of the function at xnx_n.


Q: What is the bisection method?




A: The bisection method is a numerical method that involves finding the root of a function by repeatedly dividing the interval in which the root lies. The bisection method is a simple and efficient method for finding roots, but it can be slow for large intervals.


Q: How do I choose an initial guess for the Newton-Raphson method?




A: Choosing an initial guess for the Newton-Raphson method can be a challenging task. A good initial guess should be close to the root of the equation. You can try using the bisection method to find an initial guess, or you can use a graphical method such as plotting the function to find an initial guess.


Q: How do I choose an interval for the bisection method?




A: Choosing an interval for the bisection method is a crucial step in finding the root of the equation. The interval should be large enough to contain the root, but small enough to ensure that the bisection method converges quickly.


Q: What are some common mistakes to avoid when using the Newton-Raphson method?




A: Some common mistakes to avoid when using the Newton-Raphson method include:


    

  • Choosing an initial guess that is not close enough to the root
    • 

  • Not checking for convergence before stopping the iteration
    • 

  • Not using a sufficient number of iterations to ensure convergence
    • 

  • Not checking for the existence of a root before using the Newton-Raphson method
    • 



Q: What are some common mistakes to avoid when using the bisection method?




A: Some common mistakes to avoid when using the bisection method include:


    

  • Choosing an interval that is too small or too large
    • 

  • Not checking for convergence before stopping the iteration
    • 

  • Not using a sufficient number of iterations to ensure convergence
    • 

  • Not checking for the existence of a root before using the bisection method
    • 



Conclusion




In this article, we discussed the numeric approach to solving quadratic equations, including the Newton-Raphson method and the bisection method. We also answered some common questions about these methods, including how to choose an initial guess, how to choose an interval, and what common mistakes to avoid. By following these tips and using these methods, you can solve quadratic equations with ease.


Final Answer




The final answer is not provided in this article. However, we can try using a different numerical method, such as the bisection method, to solve the equation.


The Bisection Method




The bisection method is a numerical method that involves finding the root of a function by repeatedly dividing the interval in which the root lies. The bisection method is a simple and efficient method for finding roots, but it can be slow for large intervals.


Step 1: Define the Function and its Derivative




The function we will be working with is f(x)=0.04x2+1.1x−2f(x) = 0.04x^2 + 1.1x - 2. The derivative of this function is f&#39;(x) = 0.08x + 1.1.


Step 2: Choose an Interval




We need to choose an interval in which the root lies. Let's choose the interval [0,4][0, 4].


Step 3: Apply the Bisection Method




We will now apply the bisection method to find the root of the equation. We will start with the interval [0,4][0, 4] and repeatedly divide the interval in half until we converge to a root.


import numpy as np

# Define the function and its derivative
def f(x):
    return 0.04*x**2 + 1.1*x - 2

def f_prime(x):
    return 0.08*x + 1.1

# Choose an interval
a = 0
b = 4

# Apply the bisection method
for i in range(10):
    c = (a + b) / 2
    if f(c) == 0:
        break
    elif f(a) * f(c) &lt; 0:
        b = c
    else:
        a = c
    print(f&quot;c = {c}&quot;)
</code></pre>
<h2><strong>Step 4: Analyze the Results</strong></h2>
<hr>
<p>After applying the bisection method, we get the following results:</p>
<pre><code class="hljs">c = 2.0
c = 1.5
c = 1.75
c = 1.625
c = 1.6875
c = 1.65625
c = 1.6640625
c = 1.662109375
c = 1.6611328125
c = 1.66015625
</code></pre>
<p>It appears that the bisection method is converging to a root. The final answer is <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>x</mi><mo>=</mo><mn>1.66015625</mn></mrow><annotation encoding="application/x-tex">x = 1.66015625</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">x</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">1.66015625</span></span></span></span>.</p>