Solving $x^2-2x-5=0$ Using The Quadratic Formula Real Solutions

Hey guys! Let's dive into solving quadratic equations, specifically using the quadratic formula. Quadratic equations might seem intimidating at first, but with the right approach, they're totally manageable. In this article, we'll break down the steps to find real solutions for a quadratic equation using the quadratic formula, making sure you understand every part of the process.

Understanding Quadratic Equations

Before we jump into the formula, let's make sure we're all on the same page about what a quadratic equation is. A quadratic equation is a polynomial equation of the second degree. This basically means it has a term with $x^2$ in it. The standard form of a quadratic equation is:

$ax^2 + bx + c = 0$

Where:

• 'a', 'b', and 'c' are coefficients, which are just numbers.
• 'x' is the variable we're trying to solve for.
• 'a' cannot be zero (otherwise, it wouldn't be a quadratic equation!).

Identifying Coefficients

Okay, so how do we identify 'a', 'b', and 'c' in a given equation? Let's look at an example:

$x^2 - 2x - 5 = 0$

In this equation:

• a = 1 (because there's an implied 1 in front of $x^2$)
• b = -2
• c = -5

See? Not too scary, right? Identifying these coefficients is the first crucial step in using the quadratic formula. Once you get the hang of it, you'll be spotting them in no time!

Why the Quadratic Formula?

You might be wondering, "Why do we need a special formula? Can't we just factor or something?" Well, sometimes you can factor a quadratic equation, which is awesome and often quicker. But, the quadratic formula is a universal tool that works every time, even when factoring is tricky or impossible. It's like your trusty Swiss Army knife for solving quadratic equations.

The Mighty Quadratic Formula

Alright, drumroll please... let's unveil the star of the show: the quadratic formula! This formula gives us the solutions (also called roots or zeros) for any quadratic equation in standard form.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Whoa! It might look a bit intimidating, but let's break it down piece by piece:

• $x$ represents the solutions we're trying to find.
• The plus-minus symbol ($\pm$) means there are usually two solutions: one where you add the square root part and one where you subtract it.
• The square root part ($\sqrt{b^2 - 4ac}$) is called the discriminant. It tells us a lot about the nature of the solutions (more on that later!).
• The fraction bar means everything above it is divided by everything below it. Order of operations is key here!

Memorizing the Formula

Yeah, I know, memorizing formulas isn't the most thrilling activity. But, trust me, knowing the quadratic formula by heart will save you a ton of time and effort. There are lots of tricks for memorizing it – you could write it down repeatedly, create a catchy song, or even use flashcards. Find what works best for you!

Applying the Quadratic Formula: Step-by-Step

Okay, now for the fun part: putting the formula into action! Let's use the equation from the original problem: $x^2 - 2x - 5 = 0$

Step 1: Identify a, b, and c

We already did this, but let's recap:

• a = 1
• b = -2
• c = -5

Step 2: Plug the Values into the Formula

Now, we carefully substitute these values into the quadratic formula:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-5)}}{2(1)}$

See how we just replaced 'a', 'b', and 'c' with their corresponding numbers? Make sure you pay close attention to signs – that's where a lot of mistakes happen!

Step 3: Simplify the Expression

Time to do some arithmetic! Let's break it down step-by-step:

1. Simplify the negative of -2: $-(-2) = 2$
2. Simplify inside the square root:
   • $(-2)^2 = 4$
   • $-4(1)(-5) = 20$
   • $4 + 20 = 24$
3. Simplify the denominator: $2(1) = 2$

Now our equation looks like this:

$x = \frac{2 \pm \sqrt{24}}{2}$

Step 4: Simplify the Square Root (if possible)

Can we simplify $\sqrt{24}$? Yep! We can factor out a perfect square:

$\sqrt{24} = \sqrt{4 \cdot 6} = \sqrt{4} \cdot \sqrt{6} = 2\sqrt{6}$

So now we have:

$x = \frac{2 \pm 2\sqrt{6}}{2}$

Step 5: Final Simplification

Notice that all the terms in the numerator have a common factor of 2, and so does the denominator. Let's divide everything by 2:

$x = \frac{2(1 \pm \sqrt{6})}{2} = 1 \pm \sqrt{6}$

Step 6: Write the Two Solutions

Remember the $\pm$ sign? This means we actually have two solutions:

• $x_1 = 1 + \sqrt{6}$
• $x_2 = 1 - \sqrt{6}$

And there you have it! We've found the two real solutions to the quadratic equation.

Understanding the Discriminant

Remember that part inside the square root, $b^2 - 4ac$? That's the discriminant, and it's super useful because it tells us about the nature of the solutions before we even solve the whole equation.

• If $b^2 - 4ac > 0$ (positive): There are two distinct real solutions (like in our example).
• If $b^2 - 4ac = 0$: There is exactly one real solution (a repeated root).
• If $b^2 - 4ac < 0$ (negative): There are no real solutions; the solutions are complex numbers.

In our example, the discriminant was $(-2)^2 - 4(1)(-5) = 24$, which is positive, so we knew we'd have two real solutions.

Putting It All Together

Let's recap the steps for solving quadratic equations using the quadratic formula:

1. Identify 'a', 'b', and 'c' from the standard form equation ($ax^2 + bx + c = 0$).
2. Plug the values into the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
3. Simplify the expression carefully, following the order of operations.
4. Simplify the square root if possible.
5. Further simplify by dividing out common factors.
6. Write out the two solutions using the $\pm$ symbol.
7. (Optional) Calculate the discriminant ($b^2 - 4ac$) to understand the nature of the solutions.

Conclusion

So, guys, that's the quadratic formula in a nutshell! It might seem like a lot at first, but with practice, you'll become a pro at solving quadratic equations. Remember to take it step-by-step, pay attention to those signs, and don't be afraid to break down the problem. You've got this! Keep practicing, and soon you'll be tackling those quadratic equations like a boss.