Solving F(x) = 0 For Quadratic Function F(x) = X^2 + 4x - 45
Hey math enthusiasts! Today, we're diving into solving a quadratic equation. We'll be tackling the function f(x) = x^2 + 4x - 45 and finding out where f(x) = 0. This means we're looking for the x-values that make the function equal to zero, also known as the roots or zeros of the function. So, grab your calculators and let's get started!
Understanding Quadratic Functions
First off, let's quickly recap what a quadratic function is. A quadratic function is a polynomial function of degree two, generally written in the form f(x) = ax^2 + bx + c, where a, b, and c are constants, and a is not equal to zero. The graph of a quadratic function is a parabola, a U-shaped curve that can open upwards or downwards depending on the sign of a.
In our specific case, f(x) = x^2 + 4x - 45, we have a = 1, b = 4, and c = -45. Since a is positive, the parabola opens upwards. The points where the parabola intersects the x-axis are the solutions to the equation f(x) = 0. These points are also referred to as the roots or zeros of the quadratic function. Finding these roots is what we're aiming to do.
To find the roots of a quadratic equation, we have several methods at our disposal. Some common methods include factoring, using the quadratic formula, and completing the square. Each method has its strengths, and the best one to use often depends on the specific equation we are dealing with. For our function, f(x) = x^2 + 4x - 45, we will explore factoring and the quadratic formula to illustrate different approaches and ensure a solid understanding of the solution process. Let's jump into the methods and solve this quadratic equation!
Method 1: Factoring
One of the most straightforward methods to solve quadratic equations is by factoring. Factoring involves breaking down the quadratic expression into a product of two binomials. This method is particularly efficient when the quadratic expression can be factored easily. So, let's see if we can factor f(x) = x^2 + 4x - 45.
We're looking for two numbers that multiply to c (-45) and add up to b (4). Think of it like a puzzle: we need to find the right pieces that fit together perfectly. Let's list the factor pairs of -45:
- 1 and -45
- -1 and 45
- 3 and -15
- -3 and 15
- 5 and -9
- -5 and 9
Looking at these pairs, we can see that -5 and 9 satisfy our conditions. They multiply to -45 and add up to 4. Bingo! Now, we can rewrite the quadratic expression using these numbers:
f(x) = x^2 + 4x - 45 = (x - 5)(x + 9)
Now that we have factored the quadratic expression, we can set each factor equal to zero and solve for x:
- x - 5 = 0 => x = 5
- x + 9 = 0 => x = -9
So, the solutions to f(x) = 0 are x = 5 and x = -9. This means that the parabola intersects the x-axis at these two points. Factoring is a neat way to solve quadratics when you can spot those key number pairs!
Method 2: Quadratic Formula
When factoring isn't so straightforward, the quadratic formula is our trusty backup. This formula can solve any quadratic equation, no matter how messy it looks. It's a bit like a universal key that unlocks all quadratic puzzles. The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
Remember our function, f(x) = x^2 + 4x - 45? We identified a = 1, b = 4, and c = -45. Let's plug these values into the quadratic formula and see what happens:
x = (-4 ± √(4^2 - 4 * 1 * -45)) / (2 * 1)
First, we simplify the expression under the square root:
x = (-4 ± √(16 + 180)) / 2 x = (-4 ± √196) / 2
Now, we find the square root of 196, which is 14:
x = (-4 ± 14) / 2
This gives us two possible solutions:
- x = (-4 + 14) / 2 = 10 / 2 = 5
- x = (-4 - 14) / 2 = -18 / 2 = -9
As you can see, we arrived at the same solutions as we did with factoring: x = 5 and x = -9. The quadratic formula is super reliable, especially when factoring seems like a headache. It's a great tool to have in your math toolkit!
Verification of Solutions
To ensure our solutions are correct, we can plug them back into the original function, f(x) = x^2 + 4x - 45, and check if the result is zero. This step is like the final checkmark on our math assignment – it gives us that extra bit of confidence that we've nailed it.
Let's start with x = 5:
f(5) = (5)^2 + 4(5) - 45 f(5) = 25 + 20 - 45 f(5) = 45 - 45 f(5) = 0
Great! x = 5 works perfectly. Now, let's check x = -9:
f(-9) = (-9)^2 + 4(-9) - 45 f(-9) = 81 - 36 - 45 f(-9) = 81 - 81 f(-9) = 0
Awesome! Both solutions, x = 5 and x = -9, make the function equal to zero. This confirms that our solutions are indeed correct. Plugging the solutions back into the original equation is a solid way to double-check your work and make sure you're on the right track. It's like having a built-in error detector!
Conclusion
Alright, guys, we've successfully solved for f(x) = 0 in the quadratic function f(x) = x^2 + 4x - 45! We explored two powerful methods: factoring and the quadratic formula. Factoring helped us break down the equation into simpler parts, while the quadratic formula provided a foolproof way to find the solutions, no matter how complex the equation might seem. We found that the solutions are x = 5 and x = -9. To be extra sure, we verified these solutions by plugging them back into the original function, and they both checked out!
Understanding how to solve quadratic equations is super important in math, and it pops up in all sorts of real-world scenarios, from physics to engineering. So, keep practicing these methods, and you'll become a quadratic equation-solving pro in no time! Keep up the great work, and remember, math can be fun when you break it down step by step. Until next time, happy solving!