Factoring Quadratics: Which Has One Solution?

When we talk about solving quadratic equations, one of the most fundamental methods we learn is factoring. It's like unlocking a puzzle, where we break down a complex expression into simpler parts. Today, we're going to dive into a specific type of factoring problem: identifying which quadratic equation, from a given set, has only one distinct solution. This might sound a bit tricky, but with a little bit of understanding and practice, you'll be able to spot them with ease! Let's break down what this means and how to approach it. A quadratic equation is generally represented in the form $ax^2 + bx + c = 0$, where 'a', 'b', and 'c' are constants, and 'a' is not zero. The solutions to these equations are also known as roots. When we talk about 'distinct solutions', we mean unique values for 'x' that satisfy the equation. For example, if an equation has solutions x=2 and x=3, those are two distinct solutions. However, sometimes, a quadratic equation might only have one value of 'x' that makes it true. This happens when the two roots are actually the same number. Think of it like getting the same answer twice – it's still just one unique answer. In the context of factoring, an equation having only one distinct solution usually means that the factored form of the quadratic is a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored into $(px + q)^2$ or $(px - q)^2$. When you set this factored form equal to zero, like $(px + q)^2 = 0$, taking the square root of both sides gives you $px + q = 0$, which then leads to a single solution for x. This is different from other factorable quadratics where you might get two different factors, like $(x-r1)(x-r2) = 0$, which would yield two distinct solutions, $x=r1$ and $x=r2$, unless $r1=r2$. So, our goal is to examine each of the given equations and see which one fits this perfect square trinomial pattern, leading to that single, solitary solution. We'll be using the power of factoring to uncover this unique characteristic. Let's get started on unraveling these mathematical mysteries!

Understanding Distinct Solutions in Quadratic Equations

To truly grasp which quadratic equation has only one distinct solution, we need to delve deeper into what that phrase means in the world of mathematics. Remember, a quadratic equation is an equation of the form $ax^2 + bx + c = 0$. The solutions, or roots, are the values of $x$ that make this equation true. When we talk about 'distinct' solutions, we're referring to unique values. For instance, the equation $x^2 - 5x + 6 = 0$ can be factored into $(x-2)(x-3) = 0$. This gives us two distinct solutions: $x=2$ and $x=3$. Both 2 and 3 are different from each other, hence they are distinct. Now, consider an equation like $x^2 - 4x + 4 = 0$. If we factor this, we get $(x-2)(x-2) = 0$, which can be written more compactly as $(x-2)^2 = 0$. If we set this equal to zero, we have $(x-2)^2 = 0$. To solve for $x$, we can take the square root of both sides: $\sqrt{(x-2)^2} = \sqrt{0}$. This simplifies to $x-2 = 0$. Adding 2 to both sides gives us $x=2$. In this case, even though the factored form appears to have two factors of $(x-2)$, both factors lead to the same solution, $x=2$. Therefore, this equation has only one distinct solution. This is a crucial concept because it highlights that the number of factors we see when factoring doesn't always directly translate to the number of unique answers. The key here is that when a quadratic equation has only one distinct solution, its factored form is what we call a perfect square trinomial. A perfect square trinomial is a quadratic expression that results from squaring a binomial. For example, $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$. Notice how in these expansions, the middle term is twice the product of the terms being squared, and the last term is the square of the second term. Conversely, when we encounter a trinomial of this form, we know it can be factored back into a squared binomial. So, when we're presented with a list of quadratic equations and asked to find the one with a single distinct solution, our primary strategy is to look for the equation that is a perfect square trinomial. This means checking if the quadratic expression can be expressed as $(x+k)^2$ or $(x-k)^2$ for some constant $k$. If it can, then setting it to zero will inevitably lead to only one unique value for $x$. This understanding is fundamental to mastering quadratic equations and their solutions.

Factoring Each Equation: The Key to Finding the Single Solution

Now, let's put our knowledge into practice and factor each of the given quadratic equations to determine which one yields only one distinct solution. Remember, our target is the equation that, when factored, results in a perfect square trinomial. We'll go through each option systematically, applying the principles of factoring.

Option A: $x^2+4 x+4=0$

Let's start with the first equation: $x^2+4x+4=0$. We need to find two numbers that multiply to 4 (the constant term) and add up to 4 (the coefficient of the x term). If we think about the factors of 4, we have (1, 4) and (2, 2). Let's check the sums: $1+4=5$ and $2+2=4$. Aha! The pair (2, 2) works. This means we can factor the quadratic as $(x+2)(x+2) = 0$. This can be written more compactly as $(x+2)^2 = 0$. Now, to find the solution, we set the factor equal to zero: $x+2 = 0$. Solving for $x$, we get $x = -2$. Since both factors are identical, there is only one distinct solution, which is $x=-2$. This equation fits our criteria for having a single, unique solution.

Option B: $x^2-4=0$

Next, we examine $x^2-4=0$. This equation is a bit different from the others because it's a difference of squares. The difference of squares pattern is $a^2 - b^2 = (a-b)(a+b)$. In our case, $x^2$ is $a^2$ (so $a=x$) and 4 is $b^2$ (so $b=2$). Applying the difference of squares formula, we can factor this equation as $(x-2)(x+2) = 0$. Now, we set each factor equal to zero to find the solutions:

• $x-2 = 0

  This gives us $x = 2$.

• $x+2 = 0

  This gives us $x = -2$.

As you can see, this equation has two distinct solutions: $x=2$ and $x=-2$. Therefore, this is not the equation we are looking for.

Option C: $x^2+5 x+4=0$

Moving on to $x^2+5x+4=0$. We need to find two numbers that multiply to 4 and add up to 5. Let's list the pairs of factors for 4: (1, 4) and (2, 2). Now, let's check their sums: $1+4=5$ and $2+2=4$. The pair (1, 4) works perfectly for us! So, we can factor this quadratic equation as $(x+1)(x+4) = 0$. Setting each factor to zero, we get:

• $x+1 = 0

  This yields $x = -1$.

• $x+4 = 0

  This yields $x = -4$.

Again, we have found two distinct solutions: $x=-1$ and $x=-4$. This equation does not have only one distinct solution.

Option D: $x^2-5 x+4=0$

Finally, let's look at the last equation: $x^2-5x+4=0$. We are searching for two numbers that multiply to 4 and add up to -5. Let's consider the factors of 4. We have (1, 4) and (2, 2). Since the product is positive (4) and the sum is negative (-5), both numbers must be negative. So, we consider (-1, -4) and (-2, -2). Let's check their sums: $(-1) + (-4) = -5$ and $(-2) + (-2) = -4$. The pair (-1, -4) fits our requirements. Therefore, we can factor this equation as $(x-1)(x-4) = 0$. Setting each factor to zero gives us:

• $x-1 = 0

  This leads to $x = 1$.

• $x-4 = 0

  This leads to $x = 4$.

This equation, like options B and C, also results in two distinct solutions: $x=1$ and $x=4$. So, this is also not the equation we are looking for.

Conclusion: Identifying the Equation with a Single Solution

After carefully factoring each of the provided quadratic equations, we have found our answer. The core concept we utilized was the identification of a perfect square trinomial, which is the hallmark of a quadratic equation possessing only one distinct solution. Let's recap our findings:

• Option A ($x^2+4 x+4=0$) factored into $(x+2)^2=0$, leading to the single distinct solution $x=-2$.
• Option B ($x^2-4=0$) factored into $(x-2)(x+2)=0$, yielding two distinct solutions, $x=2$ and $x=-2$.
• Option C ($x^2+5 x+4=0$) factored into $(x+1)(x+4)=0$, giving two distinct solutions, $x=-1$ and $x=-4$.
• Option D ($x^2-5 x+4=0$) factored into $(x-1)(x-4)=0$, resulting in two distinct solutions, $x=1$ and $x=4$.

Based on this analysis, it is clear that equation A, $x^2+4x+4=0$, is the one that has only one distinct solution. This is because it is a perfect square trinomial, meaning it can be written as the square of a binomial. When such an expression is set to zero, the binomial itself must be zero, resulting in a single value for the variable.

Mastering the art of factoring quadratics is a vital skill in algebra. It not only helps in solving equations but also in understanding the nature of their roots. Remember, the presence of a perfect square trinomial is your biggest clue when searching for an equation with just one unique solution. Keep practicing, and you'll become a pro at spotting these special cases!

For further exploration into quadratic equations and algebraic techniques, you can visit Khan Academy for excellent resources and practice problems. You might also find the explanations on Math is Fun very helpful for understanding mathematical concepts in an accessible way.