Consider a quadratic equation \(a x^{2}+b x+c=0\) that has real solutions. a) Write a program that takes as input the three numbers, \(a, b\), and \(c\), and prints out the two solutions using the standard formula $$ x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} . $$ Use your program to compute the solutions of \(0.001 x^{2}+1000 x+0.001=0\). b) There is another way to write the solutions to a quadratic equation. Multiplying top and bottom of the solution above by \(-b \mp \sqrt{b^{2}-4 a c}\), show that the solutions can also be written as $$ x=\frac{2 c}{-b \mp \sqrt{b^{2}-4 a c}} . $$ Add further lines to your program to print these values in addition to the earlier ones and again use the program to solve \(0.001 x^{2}+1000 x+0.001=0\). What do you see? How do you explain it? c) Using what you have learned, write a new program that calculates both roots of a quadratic equation accurately in all cases. This is a good example of how computers don't always work the way you expect them to. If you simply apply the standard formula for the quadratic equation, the computer will sometimes get the wrong answer. In practice the method you have worked out here is the correct way to solve a quadratic equation on a computer, even though it's more complicated than the standard formula. If you were writing a program that involved solving many quadratic equations this method might be a good candidate for a userdefined function: you could put the details of the solution method inside a function to save yourself the trouble of going through it step by step every time you have a new equation to solve.

Step by step solution

01

Define the problem

Consider the quadratic equation given by \(a x^{2}+b x+c=0\) that requires solving for x.

02

Standard quadratic formula

The solutions for a quadratic equation can be found using the formula \(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\). We will write a program to compute these values.

03

Write the program for the standard formula

Using the programming language of your choice, write a program that takes input values for \(a, b,\) and \c\, and computes the two roots:```pythonimport matha = 0.001b = 1000c = 0.001d = b**2 - 4*a*cx1 = (-b + math.sqrt(d)) / (2*a)x2 = (-b - math.sqrt(d)) / (2*a)print('Standard formula solutions:')print('x1 =', x1)print('x2 =', x2)```

04

Compute the first set of solutions

Run the program to get the solutions for \(0.001 x^{2}+1000 x+0.001=0\). You will get the solutions for x using the standard formula.

05

Alternative formula derivation

To derive the alternative formula, multiply both numerator and denominator by \(-b \mp \sqrt{b^{2}-4 a c}\):\[ x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \times \frac{-b \mp \sqrt{b^{2}-4 a c}}{-b \mp \sqrt{b^{2}-4 a c}} = \frac{2c} {-b \mp \sqrt{b^{2}-4 a c}}\]

06

Write the program for the alternative formula

Add additional lines to your program to implement the alternative formula:```pythonx1_alt = (2*c) / (-b + math.sqrt(d))x2_alt = (2*c) / (-b - math.sqrt(d))print('Alternative formula solutions:')print('x1_alt =', x1_alt)print('x2_alt =', x2_alt)```

07

Compute the second set of solutions

Run the updated program to get the new solutions for \(0.001 x^{2}+1000 x+0.001=0\) using the alternative formula.

08

Compare solutions

Compare the results from both formulas. You will notice that one set of solutions might be more accurate due to how numerical precision affects the calculations.

09

Write a new program for accurate computation

Based on the observed results, write a new program that selects the most accurate formula to compute the roots:```pythondef solve_quadratic(a, b, c): d = b**2 - 4*a*c if b > 0: x1 = (-b - math.sqrt(d)) / (2*a) x2 = (2*c) / (-b - math.sqrt(d)) else: x1 = (-b + math.sqrt(d)) / (2*a) x2 = (2*c) / (-b + math.sqrt(d)) return x1, x2x1, x2 = solve_quadratic(0.001, 1000, 0.001)print('Accurate solutions:')print('x1 =', x1)print('x2 =', x2)```

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Quadratic Formula

The standard quadratic formula is a powerful tool for solving quadratic equations, which take the general form \(a x^{2}+b x+c=0\). The formula to find the roots (solutions) is given by:
\(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\).
This formula directly provides the two possible solutions for the quadratic equation. Here is how it works:

• Calculate the discriminant, \(D=b^{2}-4ac\).
• If \(D < 0\), the equation has no real solutions.
• If \(D = 0\), there is one real solution: \(x=-\frac{b}{2a}\).
• If \(D > 0\), there are two real solutions: \(x=\frac{-b + \sqrt{D}}{2 a}\) and \(x=\frac{-b - \sqrt{D}}{2 a}\).

In practice, implementing the standard quadratic formula in a programming language can help compute these values efficiently. For instance in Python:
```python import math a = 0.001 b = 1000 c = 0.001 d = b**2 - 4*a*c x1 = (-b + math.sqrt(d)) / (2*a) x2 = (-b - math.sqrt(d)) / (2*a) print('Standard formula solutions:') print('x1 =', x1) print('x2 =', x2) ``` Executing the above program for the quadratic equation \(0.001 x^{2}+1000 x+0.001=0\) provides the roots using the standard formula.

Alternative Quadratic Formula

Sometimes, the standard quadratic formula might lead to precision errors, especially in cases where the coefficient of the quadratic term, \(a\), is very small compared to the linear term, \(b\).
An alternative form of the quadratic formula can help mitigate such precision issues. This formula is derived by multiplying the numerator and denominator of the standard formula by \(-b \mp \sqrt{b^{2}-4 a c}\) and is given as:
\(x=\frac{2c}{-b \mp \sqrt{b^{2}-4 a c}}\).
This alternative formula can be particularly useful when one of the roots is much larger than the other, improving numerical stability. To implement this in code:
```python x1_alt = (2*c) / (-b + math.sqrt(d)) x2_alt = (2*c) / (-b - math.sqrt(d)) print('Alternative formula solutions:') print('x1_alt =', x1_alt) print('x2_alt =', x2_alt) ```
When applied to the same equation, \(0.001 x^{2}+1000 x+0.001=0\), using the alternative formula might yield more precise results as it minimizes the potential for loss of significance.

Numerical Precision

In computational mathematics, numerical precision is a critical concept as it impacts the accuracy of results when performing floating-point arithmetic. The key here is to understand that computers have a finite precision, so certain operations might lose accuracy.
Here's how you can ensure numerical precision while solving quadratic equations:

• Use an appropriate formula based on the relative sizes of the coefficients \(a, b,\) and \(c\).
• Avoid subtracting nearly equal numbers, as this can lead to significant precision loss.
• Choose numerical methods that maintain precision across all steps of computation.

For instance, the standard quadratic formula is easy to implement but may lead to precision issues for certain coefficients. The alternative quadratic formula helps maintain precision better. To maximize accuracy, here is a combined approach in code:
```python def solve_quadratic(a, b, c): d = b**2 - 4*a*c if b > 0: x1 = (-b - math.sqrt(d)) / (2*a) x2 = (2*c) / (-b - math.sqrt(d)) else: x1 = (-b + math.sqrt(d)) / (2*a) x2 = (2*c) / (-b + math.sqrt(d)) return x1, x2 x1, x2 = solve_quadratic(0.001, 1000, 0.001) print('Accurate solutions:') print('x1 =', x1) print('x2 =', x2) ```
By selecting the most appropriate formula based on the coefficients, you ensure that your calculations are as precise as possible, minimizing errors due to numerical precision.