Solving quadratic equations with continued fractions
A quadratic equation is like a puzzle where you have to find the values of some letters (usually x) that make the equation true. Let's say we have an equation like this:
4x^2 + 6x - 2 = 0
To solve this puzzle, we need to use something called continued fractions. A fraction is when you have a number on the top and a number on the bottom, like 2/3. A continued fraction is when you keep dividing the bottom number into smaller and smaller pieces, like this:
2 / (3 + 1/(2 + 1/3))
This continued fraction is equal to the number 2.2 repeating, or 2.6666666... The important thing to remember is that continued fractions can help us solve quadratic equations.
To use continued fractions to solve a quadratic equation, we need to find the roots of the equation. The roots are the values of x that make the equation equal to zero. There are different methods for finding the roots, but we'll use the continued fractions method.
First, we need to rewrite the quadratic equation in a specific form:
ax^2 + bx + c = 0
We already have our example equation in this form. Now, we need to find a continued fraction that solves this equation.
The continued fraction that we're looking for has a specific pattern. It looks like this:
x = k + 1 / (a1 + 1 / (a2 + 1 / (a3 + ...
The k value is the integer value that makes the first term of the continued fraction equal to an integer. In our example equation, we can use k = 0. We don't need to worry too much about why we use this value - just remember that it's important to find the continued fraction.
The a1, a2, a3... values are coefficients that depend on the quadratic equation we're trying to solve. We can find these values using the following formula:
a1 = -b / a
a2 = -c / (b + a1 * a)
a3 = -b / (c + a2 * (b + a1 * a))
...
and so on.
Using this formula, we can find the values of a1, a2, a3... and plug them into our continued fraction equation:
x = 0 + 1 / (a1 + 1 / (a2 + 1 / (a3 + ...
This continued fraction will give us the roots of our quadratic equation. We can find the roots by calculating the value of the continued fraction. If the continued fraction converges - that is, if the terms get smaller and smaller, and eventually the continued fraction stops changing - then that value is one of the roots of the quadratic equation.
So, let's try this on our example equation:
a = 4
b = 6
c = -2
a1 = -6 / 4 = -3/2
a2 = -(-2) / (6 - 3/2 * 4) = 1/4
a3 = -6 / (-2 + 1/4 * (6 - 3/2 * 4)) = -2
We can plug these values into the continued fraction equation:
x = 0 + 1 / (-3/2 + 1 / (1/4 + 1 / (-2 + ...
We can keep calculating more terms of this continued fraction to get a better approximation of the root. Eventually, we'll notice that the terms start getting smaller and smaller, and the continued fraction stops changing. This means that we've found one of the roots of our quadratic equation:
x = 0.2928932... (approximately)
We can also use the quadratic formula to check our answer:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
x = (-6 ± sqrt(6^2 - 4*4*(-2))) / 8
x = (-6 ± sqrt(52)) / 8
x = (-6 ± 2sqrt(13)) / 8
x = -0.2928932... or 0.2928932...
Our continued fraction method gave us the positive root, which matches one of the roots we found using the quadratic formula. This shows that continued fractions can be a useful tool for solving quadratic equations.
4x^2 + 6x - 2 = 0
To solve this puzzle, we need to use something called continued fractions. A fraction is when you have a number on the top and a number on the bottom, like 2/3. A continued fraction is when you keep dividing the bottom number into smaller and smaller pieces, like this:
2 / (3 + 1/(2 + 1/3))
This continued fraction is equal to the number 2.2 repeating, or 2.6666666... The important thing to remember is that continued fractions can help us solve quadratic equations.
To use continued fractions to solve a quadratic equation, we need to find the roots of the equation. The roots are the values of x that make the equation equal to zero. There are different methods for finding the roots, but we'll use the continued fractions method.
First, we need to rewrite the quadratic equation in a specific form:
ax^2 + bx + c = 0
We already have our example equation in this form. Now, we need to find a continued fraction that solves this equation.
The continued fraction that we're looking for has a specific pattern. It looks like this:
x = k + 1 / (a1 + 1 / (a2 + 1 / (a3 + ...
The k value is the integer value that makes the first term of the continued fraction equal to an integer. In our example equation, we can use k = 0. We don't need to worry too much about why we use this value - just remember that it's important to find the continued fraction.
The a1, a2, a3... values are coefficients that depend on the quadratic equation we're trying to solve. We can find these values using the following formula:
a1 = -b / a
a2 = -c / (b + a1 * a)
a3 = -b / (c + a2 * (b + a1 * a))
...
and so on.
Using this formula, we can find the values of a1, a2, a3... and plug them into our continued fraction equation:
x = 0 + 1 / (a1 + 1 / (a2 + 1 / (a3 + ...
This continued fraction will give us the roots of our quadratic equation. We can find the roots by calculating the value of the continued fraction. If the continued fraction converges - that is, if the terms get smaller and smaller, and eventually the continued fraction stops changing - then that value is one of the roots of the quadratic equation.
So, let's try this on our example equation:
a = 4
b = 6
c = -2
a1 = -6 / 4 = -3/2
a2 = -(-2) / (6 - 3/2 * 4) = 1/4
a3 = -6 / (-2 + 1/4 * (6 - 3/2 * 4)) = -2
We can plug these values into the continued fraction equation:
x = 0 + 1 / (-3/2 + 1 / (1/4 + 1 / (-2 + ...
We can keep calculating more terms of this continued fraction to get a better approximation of the root. Eventually, we'll notice that the terms start getting smaller and smaller, and the continued fraction stops changing. This means that we've found one of the roots of our quadratic equation:
x = 0.2928932... (approximately)
We can also use the quadratic formula to check our answer:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
x = (-6 ± sqrt(6^2 - 4*4*(-2))) / 8
x = (-6 ± sqrt(52)) / 8
x = (-6 ± 2sqrt(13)) / 8
x = -0.2928932... or 0.2928932...
Our continued fraction method gave us the positive root, which matches one of the roots we found using the quadratic formula. This shows that continued fractions can be a useful tool for solving quadratic equations.
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