Real Solutions Of $25x^2 + 60x + 36 = 0$: How Many?

Dec 3, 2025 by Andrew McMorgan 52 views

Hey guys! Ever stumbled upon a quadratic equation and wondered how many real solutions it actually has? Well, today we're diving deep into a specific one: $25x^2 + 60x + 36 = 0$ . Don't worry, we'll break it down step-by-step so it's super easy to understand. Whether you're a math whiz or just trying to brush up on your algebra, this guide is for you. So, let's get started and figure out how many real solutions this equation holds!

Understanding Quadratic Equations

Before we jump into our specific equation, let's quickly recap what quadratic equations are all about. Quadratic equations are polynomial equations of the second degree. This means the highest power of the variable (usually x) is 2. The standard form of a quadratic equation is expressed as:

$ax^2 + bx + c = 0$

Where:

a, b, and c are coefficients, which are constants. a cannot be zero, otherwise, it would be a linear equation.
x is the variable we are trying to solve for.

The solutions to a quadratic equation are also known as the roots or zeros. These are the values of x that make the equation true. A quadratic equation can have up to two solutions because of the highest power being 2.

Now, when we talk about real solutions, we're referring to solutions that are real numbers—not imaginary numbers. This distinction is crucial because quadratic equations can sometimes have complex solutions, which involve the imaginary unit i (where $i^2 = -1$ ). To determine the nature and number of solutions, we often turn to the discriminant.

The Role of the Discriminant

So, how do we figure out how many real solutions a quadratic equation has? That's where the discriminant comes in! The discriminant is a part of the quadratic formula, and it gives us a ton of insight without actually solving the whole equation. The discriminant is denoted by the Greek letter delta ( $\Delta$ ) and is calculated using the formula:

$\Delta = b^2 - 4ac$

This simple expression is our key to unlocking the mystery of the solutions. By evaluating the discriminant, we can determine whether the quadratic equation has two distinct real solutions, one real solution (a repeated root), or no real solutions (two complex solutions).

Here's a quick rundown of how the discriminant helps us:

If $\Delta > 0$ : The equation has two distinct real solutions.
If $\Delta = 0$ : The equation has exactly one real solution (a repeated root).
If $\Delta < 0$ : The equation has no real solutions (two complex solutions).

Understanding the discriminant is absolutely essential for quickly determining the nature of the solutions without going through the entire process of solving the quadratic equation. It's a neat little trick that saves us a lot of time and effort!

Analyzing the Given Equation: $25x^2 + 60x + 36 = 0$

Okay, now that we've covered the basics, let's get back to our main problem: the quadratic equation $25x^2 + 60x + 36 = 0$ . Our mission is to figure out how many real solutions this equation has. To do that, we're going to use the discriminant we just talked about.

First things first, we need to identify the coefficients a, b, and c in our equation. Comparing $25x^2 + 60x + 36 = 0$ to the standard form $ax^2 + bx + c = 0$ , we can easily see that:

$a = 25$
$b = 60$
$c = 36$

Got those values down? Great! Now we're ready to plug them into the discriminant formula:

$\Delta = b^2 - 4ac$

Substitute the values:

$\Delta = (60)^2 - 4(25)(36)$

Let's break this down step-by-step:

Calculate $60^2$ : $60^2 = 3600$
Calculate $4(25)(36)$ : $4 * 25 = 100$ , and $100 * 36 = 3600$
Now, subtract: $\Delta = 3600 - 3600$

So, what do we get?

$\Delta = 0$

Determining the Number of Real Solutions

Alright, we've calculated the discriminant, and we found that $\Delta = 0$ . Remember what we said earlier about what the discriminant tells us? If the discriminant is equal to zero, it means the quadratic equation has exactly one real solution—also known as a repeated root.

This is a super important point: when the discriminant is zero, the quadratic equation's graph touches the x-axis at exactly one point. This point represents the single real solution. So, in our case, the equation $25x^2 + 60x + 36 = 0$ has one real solution.

To solidify our understanding, let’s quickly recap the key takeaways:

We identified the coefficients a, b, and c from the quadratic equation.
We calculated the discriminant using the formula $\Delta = b^2 - 4ac$ .
We found that $\Delta = 0$ , which indicates that the equation has one real solution.

So, there you have it! We've successfully determined the number of real solutions for the given quadratic equation. But just for kicks, let's actually find that solution, shall we?

Finding the Real Solution

Okay, so we know that our equation, $25x^2 + 60x + 36 = 0$ , has one real solution. But what is that solution? To find it, we can use the quadratic formula. Now, the quadratic formula might look a little intimidating at first, but trust me, it's not as scary as it seems. It’s given by:

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

Notice anything familiar in that formula? Yep, that's right! The expression inside the square root is none other than our trusty discriminant, $\Delta = b^2 - 4ac$ . Since we already know that $\Delta = 0$ for our equation, this simplifies things quite a bit.

Let's plug in our values for a, b, and c into the quadratic formula:

$x = \frac{-60 \pm \sqrt{0}}{2(25)}$

Since the square root of 0 is just 0, the formula simplifies further:

$x = \frac{-60 \pm 0}{50}$

This means we have:

$x = \frac{-60}{50}$

Now, let's simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 10:

$x = \frac{-6}{5}$

So, our single real solution is $x = -\frac{6}{5}$ .

Verification

Just to be absolutely sure, let's plug this value back into the original equation to verify it. We want to check if $25(-\frac{6}{5})^2 + 60(-\frac{6}{5}) + 36 = 0$ .

First, let's square $-\frac{6}{5}$ : $(-\frac{6}{5})^2 = \frac{36}{25}$
Now, multiply by 25: $25 * \frac{36}{25} = 36$
Next, multiply $60 * -\frac{6}{5}$ : $60 * -\frac{6}{5} = -72$
So, our equation becomes: $36 - 72 + 36$
Finally, let's add them up: $36 - 72 + 36 = 0$

Voila! It checks out. This confirms that $x = -\frac{6}{5}$ is indeed the single real solution to our equation.

Alternative Method: Factoring

Now, just to show you there's more than one way to solve a quadratic equation, let's take a look at another method: factoring. Factoring is a super handy technique that can sometimes be quicker than using the quadratic formula, especially when the equation is factorable.

Our equation is $25x^2 + 60x + 36 = 0$ . Notice anything special about these coefficients? They're all perfect squares or multiples of perfect squares! This hints that we might be able to factor the quadratic expression into a perfect square trinomial.

Let’s try to rewrite the equation. We can express $25x^2$ as $(5x)^2$ and $36$ as $(6)^2$ . The middle term, $60x$ , can be written as $2 * (5x) * (6)$ . Does this look familiar?

It fits the pattern of a perfect square trinomial: $(a + b)^2 = a^2 + 2ab + b^2$ .

In our case, $a = 5x$ and $b = 6$ . So, we can rewrite our equation as:

$(5x + 6)^2 = 0$

See how much simpler that looks? Now, to solve for x, we just need to take the square root of both sides:

$\sqrt{(5x + 6)^2} = \sqrt{0}$

$5x + 6 = 0$

Now, isolate x:

$5x = -6$

$x = -\frac{6}{5}$

Boom! We arrived at the same solution, $x = -\frac{6}{5}$ , using factoring. This just goes to show that understanding different methods for solving quadratic equations can be super beneficial. Factoring is a fantastic shortcut when you spot those perfect square patterns.

Conclusion

So, guys, we've reached the end of our quadratic equation adventure! We set out to determine the number of real solutions for the equation $25x^2 + 60x + 36 = 0$ , and we did just that. We learned about quadratic equations, the crucial role of the discriminant, and even explored two different methods for finding the solution.

Here’s a quick recap of what we covered:

Understanding Quadratic Equations: We revisited the standard form ( $ax^2 + bx + c = 0$ ) and the concept of real solutions.
The Discriminant: We learned how to use the discriminant ( $\Delta = b^2 - 4ac$ ) to determine the number of real solutions.
Analyzing $25x^2 + 60x + 36 = 0$ : We identified a, b, and c, calculated the discriminant ( $\Delta = 0$ ), and concluded that the equation has one real solution.
Finding the Real Solution: We used the quadratic formula to find that single real solution, $x = -\frac{6}{5}$ , and verified our answer.
Alternative Method: Factoring: We demonstrated how factoring could also lead us to the same solution, highlighting the versatility of different problem-solving techniques.

We discovered that the equation $25x^2 + 60x + 36 = 0$ has exactly one real solution, which is $x = -\frac{6}{5}$ . Whether you prefer using the discriminant and the quadratic formula or spotting patterns for factoring, you now have a solid understanding of how to tackle similar problems.

Remember, math isn't just about numbers and formulas; it's about problem-solving and critical thinking. Keep practicing, keep exploring, and you'll become a math whiz in no time! Until next time, keep those equations balanced and those solutions real!