Find The Smaller Solution To $x^2 - 5x + 3 = 0$

by Alex Johnson 48 views

When you're faced with a quadratic equation, like the one we have here, x2βˆ’5x+3=0x^2 - 5x + 3 = 0, finding its solutions can sometimes feel like a puzzle. But don't worry, it's a puzzle we can solve with the right tools! Today, we're going to dive deep into how to find the solutions to this specific quadratic equation and identify the smaller one. We'll be using the quadratic formula, a powerful tool that helps us solve any equation of the form ax2+bx+c=0ax^2 + bx + c = 0. This formula is your best friend when factoring becomes tricky or impossible. So, let's get started on unraveling the mystery behind these solutions and pick out the smallest one from the options provided.

Understanding the Quadratic Formula

The quadratic formula is a general solution for quadratic equations. For any equation in the standard form ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are coefficients and aa is not equal to zero, the solutions for xx are given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This formula might look a bit intimidating at first, but it's incredibly reliable. The part under the square root, b2βˆ’4acb^2 - 4ac, is called the discriminant. The discriminant tells us about the nature of the solutions: if it's positive, we have two distinct real solutions; if it's zero, we have one real solution (a repeated root); and if it's negative, we have two complex solutions. In our case, the equation is x2βˆ’5x+3=0x^2 - 5x + 3 = 0. Let's identify the coefficients aa, bb, and cc. Here, a=1a = 1 (the coefficient of x2x^2), b=βˆ’5b = -5 (the coefficient of xx), and c=3c = 3 (the constant term). Now, we can plug these values into the quadratic formula to find the solutions. It's crucial to be careful with the signs, especially when substituting negative values for bb. This formula provides a systematic way to solve these equations, ensuring that you don't miss any potential solutions and can compare them accurately to find the smaller one. The beauty of the quadratic formula lies in its universality; it works for all quadratic equations, making it an indispensable tool in algebra. We'll go through each step meticulously to ensure accuracy in our calculations and to build a solid understanding of the process involved in solving quadratic equations.

Applying the Formula to x2βˆ’5x+3=0x^2 - 5x + 3 = 0

Let's plug our values (a=1a=1, b=βˆ’5b=-5, c=3c=3) into the quadratic formula:

x=βˆ’(βˆ’5)Β±(βˆ’5)2βˆ’4(1)(3)2(1)x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(3)}}{2(1)}

First, let's simplify the terms inside the formula. The βˆ’(βˆ’5)-(-5) becomes 55. The (βˆ’5)2(-5)^2 becomes 2525. And 4(1)(3)4(1)(3) is 1212. So, the formula now looks like:

x=5Β±25βˆ’122x = \frac{5 \pm \sqrt{25 - 12}}{2}

Now, we calculate the value under the square root (the discriminant): 25βˆ’12=1325 - 12 = 13. So, we have:

x=5Β±132x = \frac{5 \pm \sqrt{13}}{2}

This gives us two distinct solutions because the discriminant (1313) is positive. The two solutions are:

  1. x1=5+132x_1 = \frac{5 + \sqrt{13}}{2}
  2. x2=5βˆ’132x_2 = \frac{5 - \sqrt{13}}{2}

These are the two roots of our quadratic equation. We've successfully applied the quadratic formula and found the expressions for both solutions. The next step, and a very important one, is to determine which of these two is the smaller solution. This involves understanding how the plus and minus signs in the formula affect the final value. When we have a term like 13\sqrt{13}, it represents a positive value. Adding it to 55 will result in a larger number than subtracting it from 55. Therefore, the solution with the minus sign will invariably be the smaller one. This is a general principle to keep in mind when comparing solutions derived from the quadratic formula. The structure of the formula itself dictates this relationship between the two roots. We are almost there in finding the answer to our problem, and this step of comparison is key to isolating the specific solution requested.

Identifying the Smaller Solution

We have our two solutions: 5+132\frac{5 + \sqrt{13}}{2} and 5βˆ’132\frac{5 - \sqrt{13}}{2}. To find the smaller solution, we need to compare these two values. Notice that both solutions share the same numerator structure (5Β±135 \pm \sqrt{13}) and the same denominator (22). The only difference lies in the sign connecting 55 and 13\sqrt{13}.

  • The first solution, 5+132\frac{5 + \sqrt{13}}{2}, involves adding 13\sqrt{13} to 55. Since 13\sqrt{13} is a positive number (approximately 3.63.6), this will result in a larger sum.
  • The second solution, 5βˆ’132\frac{5 - \sqrt{13}}{2}, involves subtracting 13\sqrt{13} from 55. Since we are subtracting a positive value from 55, the result will be smaller than if we were adding it.

Therefore, the smaller solution is the one with the minus sign: 5βˆ’132\frac{5 - \sqrt{13}}{2}. This is a direct consequence of the properties of addition and subtraction with positive numbers. When comparing two fractions with the same positive denominator, the fraction with the smaller numerator will be the smaller value. In this case, 5βˆ’135 - \sqrt{13} is clearly smaller than 5+135 + \sqrt{13}. This systematic comparison ensures we correctly identify the smaller root as requested by the problem. It's a straightforward comparison once the solutions are presented in their simplified form.

Comparing with the Given Options

Now, let's look at the options provided to see which one matches our smaller solution:

A. 5132\frac{5 \sqrt{13}}{2} B. βˆ’5132-\frac{5 \sqrt{13}}{2} C. 5βˆ’132\frac{5-\sqrt{13}}{2} D. 5+132\frac{5+\sqrt{13}}{2}

By comparing our calculated smaller solution, 5βˆ’132\frac{5 - \sqrt{13}}{2}, with the given options, we can see that Option C is the perfect match. Options A and B involve 5135\sqrt{13} in the numerator, which is a different structure entirely and doesn't arise from our application of the quadratic formula. Option D, 5+132\frac{5+\sqrt{13}}{2}, is the larger of the two solutions we found. Therefore, the correct answer, representing the smaller solution to the quadratic equation x2βˆ’5x+3=0x^2 - 5x + 3 = 0, is indeed 5βˆ’132\frac{5 - \sqrt{13}}{2}. This process of matching our result with the given choices confirms our calculations and solidifies our understanding of how to interpret the output of the quadratic formula. It’s a satisfying conclusion to have found the correct answer and to know why it is the correct answer.

Conclusion

We've successfully navigated the process of solving a quadratic equation using the quadratic formula. By carefully identifying the coefficients aa, bb, and cc from the equation x2βˆ’5x+3=0x^2 - 5x + 3 = 0, we plugged them into the formula x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. This led us to the two solutions: 5+132\frac{5 + \sqrt{13}}{2} and 5βˆ’132\frac{5 - \sqrt{13}}{2}. Through a direct comparison, we determined that 5βˆ’132\frac{5 - \sqrt{13}}{2} is the smaller solution. This result perfectly matched Option C among the choices provided. Understanding how to solve quadratic equations is a fundamental skill in mathematics, opening doors to more complex problems in algebra, calculus, and beyond. The quadratic formula is a powerful and versatile tool that guarantees a solution for any quadratic equation you encounter.

For further exploration into the fascinating world of quadratic equations and other algebraic concepts, you can visit trusted resources like Khan Academy. They offer comprehensive explanations, practice problems, and video tutorials to deepen your understanding.