Solve $m^2 - 3m - 18 = 0$ With Quadratic Formula

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Solving $m^2 - 3m - 18 = 0$ Using the Quadratic Formula

Hey guys! Today, we're going to dive into solving the quadratic equation m2βˆ’3mβˆ’18=0m^2 - 3m - 18 = 0. We'll be using the quadratic formula, a super handy tool for tackling these types of problems. Trust me, once you get the hang of it, you'll be solving quadratic equations like a pro!

Understanding the Quadratic Formula

First things first, let's talk about what the quadratic formula actually is. A quadratic equation is generally written in the form ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are coefficients. The quadratic formula provides a way to find the values of xx (or in our case, mm) that satisfy the equation. The formula itself looks like this:

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

It might look a bit intimidating at first, but don't worry! We'll break it down step by step. The Β±\pm symbol means that there are two possible solutions: one where you add the square root part and one where you subtract it. This is because quadratic equations can have up to two real solutions.

In essence, the quadratic formula is derived from the process of completing the square on the general form of a quadratic equation. Completing the square transforms the quadratic equation into a form that allows us to easily isolate the variable x. The quadratic formula is simply the generalized result of this process, providing a direct method to find the solutions without having to complete the square each time. Understanding this derivation can give you a deeper appreciation for why the formula works. The term inside 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, there are two distinct real solutions; if it's zero, there is exactly one real solution (a repeated root); and if it's negative, there are two complex solutions. This is why the quadratic formula is so powerful: it not only gives us the solutions, but also tells us what kind of solutions to expect. Mastering the quadratic formula is a fundamental skill in algebra, providing a robust method for solving a wide range of quadratic equations efficiently and accurately. Remember to practice applying the formula to various problems to solidify your understanding. With enough practice, you will be able to quickly identify the coefficients, plug them into the formula, and simplify to find the solutions. This will be invaluable for more advanced topics in mathematics and its applications.

Identifying a, b, and c

Okay, now let's identify the coefficients aa, bb, and cc in our equation, m2βˆ’3mβˆ’18=0m^2 - 3m - 18 = 0. Comparing this to the general form ax2+bx+c=0ax^2 + bx + c = 0, we can see that:

  • a=1a = 1 (since m2m^2 is the same as 1m21m^2)
  • b=βˆ’3b = -3
  • c=βˆ’18c = -18

It's super important to pay attention to the signs! A negative sign can totally change your answer, so make sure you're grabbing the correct values.

Understanding how to correctly identify aa, bb, and cc is critical because these values directly feed into the quadratic formula. If you misidentify even one of these coefficients, your entire calculation will be off, leading to incorrect solutions. So, take your time and double-check each value against the standard quadratic equation form. For instance, if the equation were 2m2+5mβˆ’7=02m^2 + 5m - 7 = 0, then aa would be 2, bb would be 5, and cc would be -7. Always include the sign! When a is 1, it's often implied and not explicitly written, like in our example m2βˆ’3mβˆ’18=0m^2 - 3m - 18 = 0. Don't forget that it's still there! Practice identifying these coefficients in a variety of quadratic equations to build your confidence. The more you practice, the faster and more accurate you'll become. Remember, precision at this stage is key to successfully solving the equation using the quadratic formula. Also, keep in mind that sometimes quadratic equations are not presented in the standard form right away. You might need to rearrange the terms to get it into the ax2+bx+c=0ax^2 + bx + c = 0 format. For example, if you have 5m2=3m+25m^2 = 3m + 2, you would need to rewrite it as 5m2βˆ’3mβˆ’2=05m^2 - 3m - 2 = 0 before identifying a=5a = 5, b=βˆ’3b = -3, and c=βˆ’2c = -2. This preliminary step is just as important as plugging the values into the formula itself.

Plugging the Values into the Formula

Now comes the fun part: plugging our values into the quadratic formula:

m=βˆ’(βˆ’3)Β±(βˆ’3)2βˆ’4(1)(βˆ’18)2(1)m = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-18)}}{2(1)}

Notice how we're substituting aa, bb, and cc into their respective places in the formula. Be extra careful with the negative signs! A double negative (like βˆ’(βˆ’3)-(-3)) becomes a positive.

When substituting the values of aa, bb, and cc into the quadratic formula, it's a good practice to use parentheses. This helps to avoid confusion, especially with negative numbers. For instance, instead of writing βˆ’32-3^2, write (βˆ’3)2(-3)^2. This makes it clear that you are squaring the entire value of -3, not just the 3. Another common mistake is forgetting to multiply 2a2a in the denominator. Make sure you carry out this multiplication correctly. After substituting, take a moment to visually inspect your work. Ensure that each value is in the correct place and that you've accounted for all the negative signs. A small error here can lead to significantly different results, so accuracy is paramount. Also, remember the order of operations (PEMDAS/BODMAS). You need to evaluate the expression inside the square root first, before you can add or subtract it from βˆ’b-b. Specifically, calculate (βˆ’3)2(-3)^2, then multiply 4(1)(βˆ’18)4(1)(-18), and finally subtract the second result from the first. By paying close attention to these details, you can minimize the chances of making errors and increase your confidence in using the quadratic formula. Each step in this process contributes to the final solution, so make sure you're methodical and precise. This will ultimately make you more proficient and comfortable with solving quadratic equations. Remember, practice makes perfect, so don't be afraid to work through multiple examples to solidify your understanding.

Simplifying the Expression

Let's simplify that beast of an equation we've got:

m=3Β±9+722m = \frac{3 \pm \sqrt{9 + 72}}{2}

m=3Β±812m = \frac{3 \pm \sqrt{81}}{2}

m=3Β±92m = \frac{3 \pm 9}{2}

See how we first dealt with the stuff inside the square root? We squared -3 to get 9, multiplied 4 * 1 * -18 to get -72, and then added 9 and 72 to get 81. Finally, we took the square root of 81, which is 9.

Simplifying the expression under the square root is a crucial step because it directly impacts the final solutions you'll obtain. Always double-check your arithmetic, especially when dealing with negative numbers and exponents. Make sure you follow the correct order of operations (PEMDAS/BODMAS) to avoid any miscalculations. After you've simplified the square root, you'll have an expression in the form of m=(AΒ±B)/Cm = (A \pm B) / C. This means you need to calculate two separate values for m: one where you add B to A and divide by C, and another where you subtract B from A and divide by C. This is because quadratic equations can have two distinct solutions. For instance, if you end up with m=(5Β±3)/2m = (5 \pm 3) / 2, you would calculate m=(5+3)/2=4m = (5 + 3) / 2 = 4 and m=(5βˆ’3)/2=1m = (5 - 3) / 2 = 1. These are your two solutions. Simplification is not just about getting the correct numerical values; it's also about making the expression as clean and manageable as possible. This reduces the likelihood of making errors in subsequent steps. Pay attention to common squares and roots to speed up the process. Knowing that the square root of 81 is 9, for example, saves you time and effort. Practice with a variety of equations to build your simplification skills and become more confident in your ability to handle complex expressions. Remember, accuracy and efficiency are key when simplifying expressions, so take your time and double-check your work.

Finding the Two Solutions

Now we split the Β±\pm into two separate equations:

m1=3+92=122=6m_1 = \frac{3 + 9}{2} = \frac{12}{2} = 6

m2=3βˆ’92=βˆ’62=βˆ’3m_2 = \frac{3 - 9}{2} = \frac{-6}{2} = -3

So, our two solutions are m=6m = 6 and m=βˆ’3m = -3.

Finding the two solutions involves carefully separating the Β±\pm sign and performing two distinct calculations. Make sure you carry out both the addition and subtraction correctly, paying close attention to the signs. It's a good idea to label your solutions, for example, as m1m_1 and m2m_2, to keep them organized. After you've found the two solutions, you can verify them by plugging them back into the original quadratic equation. If both solutions satisfy the equation, then you know you've done everything correctly. For instance, let's verify m=6m = 6 in the equation m2βˆ’3mβˆ’18=0m^2 - 3m - 18 = 0: (6)2βˆ’3(6)βˆ’18=36βˆ’18βˆ’18=0(6)^2 - 3(6) - 18 = 36 - 18 - 18 = 0. This confirms that m=6m = 6 is indeed a solution. Similarly, let's verify m=βˆ’3m = -3: (βˆ’3)2βˆ’3(βˆ’3)βˆ’18=9+9βˆ’18=0(-3)^2 - 3(-3) - 18 = 9 + 9 - 18 = 0. This confirms that m=βˆ’3m = -3 is also a solution. Verification is a valuable step because it allows you to catch any errors you might have made along the way. If a solution doesn't satisfy the original equation, then you know you need to go back and check your work. Accuracy and thoroughness are essential when finding and verifying solutions to quadratic equations. Also, remember that not all quadratic equations have two distinct real solutions. Some have one real solution (a repeated root), and others have two complex solutions. The nature of the solutions depends on the value of the discriminant (b2βˆ’4acb^2 - 4ac).

Conclusion

And there you have it! We successfully used the quadratic formula to solve the equation m2βˆ’3mβˆ’18=0m^2 - 3m - 18 = 0. Remember the steps:

  1. Identify aa, bb, and cc.
  2. Plug the values into the quadratic formula.
  3. Simplify the expression.
  4. Find the two solutions.

Keep practicing, and you'll become a quadratic equation-solving master in no time! You got this!