Solve $x^2-10x+2=0$: Find Prime $a$ And $b$

Understanding Quadratic Equations and Prime Numbers

Quadratic equations are a fundamental concept in algebra, typically expressed in the form ax² + bx + c = 0, where 'a', 'b', and 'c' are constants and 'a' is not zero. The solutions to these equations, often called roots, can reveal a great deal about the underlying mathematical relationship. In this particular problem, we're diving into the equation $x^2 - 10x + 2 = 0$. We're not just looking for any solutions; we're seeking solutions that can be expressed in a specific format: $a e ext{sqrt}(b)$. This format adds an extra layer of complexity, as both 'a' and 'b' are required to be prime numbers. Prime numbers are whole numbers greater than 1 that have only two divisors: 1 and themselves. Think of numbers like 2, 3, 5, 7, 11, and so on. The challenge here is to not only solve the quadratic equation but also to identify the specific prime numbers that fit the given solution structure. This involves understanding how to use the quadratic formula and then critically examining the results to see if they align with the prime number criteria. It's a great exercise in bridging different mathematical concepts.

Applying the Quadratic Formula

To solve the equation $x^2 - 10x + 2 = 0$, we can use the well-known quadratic formula. This formula is a lifeline for any quadratic equation that can't be easily factored. The formula states that for an equation in the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

x = rac{-b e ext{sqrt}(b^2 - 4ac)}{2a}

In our specific equation, $x^2 - 10x + 2 = 0$, we can identify the coefficients: $a = 1$, $b = -10$, and $c = 2$. Now, let's substitute these values into the quadratic formula:

x = rac{-(-10) e ext{sqrt}((-10)^2 - 4 * 1 * 2)}{2 * 1}

Let's simplify this step by step. First, the term $-b$ becomes $-(-10) = 10$. Then, inside the square root, we have $(-10)^2 = 100$. The term $4ac$ is $4 * 1 * 2 = 8$. So, the expression inside the square root becomes $100 - 8 = 92$. The denominator $2a$ is $2 * 1 = 2$.

Putting it all together, we get:

x = rac{10 e ext{sqrt}(92)}{2}

Now, we need to simplify the square root of 92. We look for the largest perfect square that divides 92. In this case, 4 is a perfect square ($2^2 = 4$), and $92 = 4 * 23$. So, $ ext{sqrt}(92) = ext{sqrt}(4 * 23) = ext{sqrt}(4) * ext{sqrt}(23) = 2 * ext{sqrt}(23)$.

Substituting this back into our solution for $x$:

x = rac{10 e 2 * ext{sqrt}(23)}{2}

Finally, we can divide both terms in the numerator by 2:

x = rac{10}{2} e rac{2 * ext{sqrt}(23)}{2}

$x = 5 e ext{sqrt}(23)$

So, the solutions to the equation $x^2 - 10x + 2 = 0$ are $5 + ext{sqrt}(23)$ and $5 - ext{sqrt}(23)$.

Identifying $a$ and $b$ as Prime Numbers

We have successfully solved the quadratic equation $x^2 - 10x + 2 = 0$, and the solutions are in the form $x = 5 e ext{sqrt}(23)$. The problem states that these solutions can be written as $a e ext{sqrt}(b)$, where a and b are prime numbers. By comparing our solutions, $5 e ext{sqrt}(23)$, with the given format $a e ext{sqrt}(b)$, we can directly identify the values of $a$ and $b$.

In this case, $a = 5$ and $b = 23$.

Now, the crucial part is to verify if these identified values are indeed prime numbers. Let's examine $a = 5$. The only positive integers that divide 5 are 1 and 5. Since 5 is greater than 1 and has only two distinct positive divisors, 5 is a prime number. This satisfies the condition for $a$.

Next, let's examine $b = 23$. The only positive integers that divide 23 are 1 and 23. Since 23 is greater than 1 and has only two distinct positive divisors, 23 is also a prime number. This satisfies the condition for $b$.

Both $a=5$ and $b=23$ meet the requirement of being prime numbers. Therefore, the solutions to the equation $x^2 - 10x + 2 = 0$ are indeed in the specified form $a e ext{sqrt}(b)$ where $a$ and $b$ are prime numbers.

Conclusion: The Prime Solutions Revealed

We embarked on a journey to solve the quadratic equation $x^2 - 10x + 2 = 0$ with a specific condition: its solutions must be expressible in the form $a e ext{sqrt}(b)$, where both $a$ and $b$ are prime numbers. By diligently applying the quadratic formula, we found the roots to be $x = 5 e ext{sqrt}(23)$. A direct comparison with the required format immediately pointed to $a = 5$ and $b = 23$. Our final step was to confirm the primality of these numbers. Indeed, 5 is prime as it's only divisible by 1 and itself, and similarly, 23 is prime, divisible only by 1 and 23. Thus, we have successfully met all the conditions of the problem.

The values of $a$ and $b$ are $a = 5$ and $b = 23$. These are the prime numbers that fit the unique structure of the solutions for the given quadratic equation. This problem beautifully illustrates how different mathematical concepts, like quadratic equations and number theory (specifically prime numbers), can intersect to create intriguing challenges and solutions.

If you're interested in exploring more about quadratic equations and their properties, a great resource is the Wolfram MathWorld.

For further understanding of prime numbers and their significance in mathematics, you might find The Prime Pages very insightful.