Solving Quadratic Equations: Step-by-Step Solutions

Hey guys! Today, we're diving into the exciting world of quadratic equations. If you've ever felt a little lost trying to solve these, don't worry – we're going to break it down step-by-step. We'll tackle equations in the form of $ax^2 + bx + c = 0$, and trust me, it's going to be a lot easier than you think. So, let's grab our pencils and notebooks and get started!

Understanding Quadratic Equations

Before we jump into solving, let's make sure we're all on the same page about what a quadratic equation actually is.

The heart of understanding quadratic equations lies in recognizing their general form: $ax^2 + bx + c = 0$. Here, a, b, and c are constants, and x is our variable. The most important part? The $x^2$ term, which makes it quadratic. If you don't have that squared term, you're dealing with a linear equation, not a quadratic one.

Think of it like this: if you're plotting a quadratic equation on a graph, you'll get a parabola – that classic U-shaped curve. Linear equations, on the other hand, give you straight lines. So, that $x^2$ term really changes the game. Knowing this general form is your first step because it helps you identify the a, b, and c values, which we'll need later for solving. Keep this in mind, and you're already halfway there!

Now, why should you care about quadratic equations? Well, they pop up everywhere in the real world! They're used in physics to describe projectile motion, in engineering to design bridges, and even in finance to model growth and decay. So, understanding these equations isn't just about acing your math test; it's about understanding the world around you.

Understanding the real-world applications makes learning quadratic equations not just an academic exercise, but a practical skill.

Methods for Solving Quadratic Equations

Okay, now we know what quadratic equations are, so let's get into how to solve them. There are a few main methods we can use, and the best one to choose often depends on the specific equation you're facing. We'll cover factoring, using the square root property, and the quadratic formula.

1. Factoring

Factoring is like the cool shortcut in the world of quadratic equations. If you can spot the factors, you can solve the equation pretty quickly. The basic idea is to rewrite the quadratic equation as a product of two binomials. Let's say you have an equation like $x^2 + 5x + 6 = 0$. You need to find two numbers that multiply to 6 (the constant term) and add up to 5 (the coefficient of the x term). In this case, those numbers are 2 and 3. So, you can factor the equation as $(x + 2)(x + 3) = 0$.

Now, here's the clever part: if the product of two things is zero, then at least one of them has to be zero. So, either $x + 2 = 0$ or $x + 3 = 0$. Solving these simple equations gives you the solutions $x = -2$ and $x = -3$. See how quick that was when the numbers play nice? However, factoring isn't always straightforward. It works best when the roots are integers, and sometimes you'll encounter equations that just don't factor easily. That's when we need our other tools.

2. Using the Square Root Property

The square root property is super handy for quadratic equations that are missing the 'bx' term – equations in the form $ax^2 + c = 0$. The goal here is to isolate the $x^2$ term on one side of the equation and then take the square root of both sides. Remember that taking the square root gives you both positive and negative solutions, which is a key point to remember.

For example, let's solve $x^2 - 9 = 0$. First, add 9 to both sides to get $x^2 = 9$. Now, take the square root of both sides: $\sqrt{x^2} = \pm\sqrt{9}$. This gives you $x = \pm 3$, so the solutions are $x = 3$ and $x = -3$. Easy peasy, right? This method is a real timesaver when you spot an equation in this simplified form. But, like factoring, it's not a universal solution. For more complex equations with the 'bx' term, we need the big guns: the quadratic formula.

3. The Quadratic Formula

Ah, the quadratic formula – the ultimate Swiss Army knife for solving quadratic equations. It might look a little intimidating at first, but it works every single time, no matter how messy the equation. If you can memorize this formula, you can solve any quadratic equation, guaranteed. The formula is:

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

Remember our general form, $ax^2 + bx + c = 0$? The a, b, and c in the formula are the same coefficients from your equation. Let's see it in action. Suppose we want to solve $2x^2 + 5x - 3 = 0$. Here, $a = 2$, $b = 5$, and $c = -3$. Plug these values into the formula:

$$x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)}$$

Simplify step by step:

$$x = \frac{-5 \pm \sqrt{25 + 24}}{4}$$

$$x = \frac{-5 \pm \sqrt{49}}{4}$$

$$x = \frac{-5 \pm 7}{4}$$

Now, split the \pm into two cases:

$$x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}$$

$$x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3$$

So, the solutions are $x = \frac{1}{2}$ and $x = -3$. The quadratic formula might involve more steps than factoring or the square root property, but it's a foolproof method. It's your best friend when you're faced with a tricky equation that doesn't seem to fit the other methods.

Solving the Equations

Alright, let's put our knowledge to the test and solve those equations you provided. We'll walk through each one, so you can see the methods in action.

a) $x^2 - 7x = 0$

For this equation, the best approach is factoring. Notice that both terms have an x in them, so we can factor out an x:

$$x(x - 7) = 0$$

Now, set each factor equal to zero:

$$x = 0$$

$$x - 7 = 0 \implies x = 7$$

So, the solutions are $x = 0$ and $x = 7$.

b) $4x^2 - 12x = 0$

Again, factoring is our friend here. We can factor out $4x$ from both terms:

$$4x(x - 3) = 0$$

Set each factor equal to zero:

$$4x = 0 \implies x = 0$$

$$x - 3 = 0 \implies x = 3$$

The solutions are $x = 0$ and $x = 3$.

c) $2x^2 - 10x = 0$

Time for more factoring! Factor out $2x$:

$$2x(x - 5) = 0$$

Set each factor equal to zero:

$$2x = 0 \implies x = 0$$

$$x - 5 = 0 \implies x = 5$$

Our solutions are $x = 0$ and $x = 5$.

d) $2x^2 - 50 = 0$

This one is perfect for the square root property. First, isolate the $x^2$ term:

$$2x^2 = 50$$

Divide by 2:

$$x^2 = 25$$

Take the square root of both sides (remembering the plus and minus):

$$x = \pm \sqrt{25}$$

So, $x = \pm 5$, which means the solutions are $x = 5$ and $x = -5$.

e) $5x^2 + 20 = 0$

Let's try the square root property again. Isolate $x^2$:

$$5x^2 = -20$$

Divide by 5:

$$x^2 = -4$$

Now, take the square root of both sides:

$$x = \pm \sqrt{-4}$$

Wait a minute! We have the square root of a negative number. That means there are no real solutions. This equation has complex solutions, but for this exercise, we'll say there are no real solutions.

f) $3x^2 + 48 = 0$

Same situation as the last one. Isolate $x^2$:

$$3x^2 = -48$$

Divide by 3:

$$x^2 = -16$$

Take the square root:

$$x = \pm \sqrt{-16}$$

Again, we have the square root of a negative number, so there are no real solutions.

g) $x^2 - 5x + 6 = 0$

This looks like a good candidate for factoring. We need two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3:

$$(x - 2)(x - 3) = 0$$

Set each factor equal to zero:

$$x - 2 = 0 \implies x = 2$$

$$x - 3 = 0 \implies x = 3$$

So, the solutions are $x = 2$ and $x = 3$.

h) $x^2 - 8x + 12 = 0$

More factoring! We need two numbers that multiply to 12 and add to -8. Those numbers are -2 and -6:

$$(x - 2)(x - 6) = 0$$

Set each factor equal to zero:

$$x - 2 = 0 \implies x = 2$$

$$x - 6 = 0 \implies x = 6$$

The solutions are $x = 2$ and $x = 6$.

i) $3x^2 - 7x + 2 = 0$

This one might be a bit trickier to factor, so let's use the quadratic formula. Here, $a = 3$, $b = -7$, and $c = 2$. Plug these into the formula:

$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(2)}}{2(3)}$$

Simplify:

$$x = \frac{7 \pm \sqrt{49 - 24}}{6}$$

$$x = \frac{7 \pm \sqrt{25}}{6}$$

$$x = \frac{7 \pm 5}{6}$$

Split into two cases:

$$x = \frac{7 + 5}{6} = \frac{12}{6} = 2$$

$$x = \frac{7 - 5}{6} = \frac{2}{6} = \frac{1}{3}$$

So, the solutions are $x = 2$ and $x = \frac{1}{3}$.

Conclusion

There you have it! We've solved a bunch of quadratic equations using factoring, the square root property, and the quadratic formula. Remember, the key is to choose the right method for the equation you're dealing with. Factoring is great when it works, the square root property is speedy for certain forms, and the quadratic formula is your reliable backup for any situation.

Keep practicing, and you'll become a quadratic equation-solving pro in no time! And remember, math isn't about memorizing formulas; it's about understanding the concepts and applying them. So, keep exploring, keep questioning, and keep learning. You've got this!