No Real Solutions: Solving $2x^2 - 4x = T$

by Omar Yusuf 43 views

Hey everyone! Let's dive into a fascinating math problem today. We're going to explore the quadratic equation 2x2−4x=t2x^2 - 4x = t, where tt is a constant. Our main goal is to figure out which value of tt would make this equation have no real solutions. Sounds intriguing, right? So, grab your thinking caps, and let's get started!

Understanding the Quadratic Equation and Its Solutions

Before we jump into the specifics of our problem, let's take a step back and make sure we're all on the same page about quadratic equations. A quadratic equation is basically a polynomial equation of the second degree. That means the highest power of the variable (in our case, x) is 2. The general form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0, where a, b, and c are constants, and a is not equal to 0. In our equation, 2x2−4x=t2x^2 - 4x = t, we can rewrite it as 2x2−4x−t=02x^2 - 4x - t = 0 to fit this general form, where a = 2, b = -4, and c = -t.

Now, what does it mean to have 'solutions' to a quadratic equation? Well, the solutions are the values of x that make the equation true. Graphically, these solutions are the points where the parabola represented by the quadratic equation intersects the x-axis. A quadratic equation can have two real solutions, one real solution (which we call a repeated root), or no real solutions. The key to figuring out how many solutions a quadratic equation has lies in something called the discriminant.

The discriminant, often denoted by the Greek letter delta (Δ), is a part of the quadratic formula that tells us about the nature of the roots. The quadratic formula is used to find the solutions of a quadratic equation and is given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The discriminant is the expression under the square root, which is b2−4acb^2 - 4ac. The value of the discriminant tells us a lot:

  • If b2−4ac>0b^2 - 4ac > 0, the equation has two distinct real solutions.
  • If b2−4ac=0b^2 - 4ac = 0, the equation has one real solution (a repeated root).
  • If b2−4ac<0b^2 - 4ac < 0, the equation has no real solutions. This is because we can't take the square root of a negative number in the real number system.

So, for our problem, we're particularly interested in the case where the discriminant is less than 0, as this is when the equation has no real solutions. Let's keep this in mind as we move forward.

Applying the Discriminant to Our Equation

Alright, now that we've brushed up on our knowledge of quadratic equations and the discriminant, let's apply this to our specific equation: 2x2−4x=t2x^2 - 4x = t. As we discussed earlier, we can rewrite this as 2x2−4x−t=02x^2 - 4x - t = 0. This allows us to identify the coefficients: a = 2, b = -4, and c = -t. Our mission is to find the value of t that makes the equation have no real solutions. That means we need to find t such that the discriminant, b2−4acb^2 - 4ac, is less than 0.

Let's plug in the values of a, b, and c into the discriminant:

b2−4ac=(−4)2−4(2)(−t)=16+8tb^2 - 4ac = (-4)^2 - 4(2)(-t) = 16 + 8t

Now, we want this discriminant to be less than 0, so we set up the inequality:

16+8t<016 + 8t < 0

To solve for t, we first subtract 16 from both sides:

8t<−168t < -16

Then, we divide both sides by 8:

t<−2t < -2

This inequality tells us that for the equation 2x2−4x=t2x^2 - 4x = t to have no real solutions, the value of t must be less than -2. This is a crucial piece of information that will help us choose the correct answer from the given options.

Evaluating the Answer Choices

Okay, we've determined that the equation 2x2−4x=t2x^2 - 4x = t has no real solutions when t is less than -2. Now, let's look at the answer choices provided and see which one fits this condition:

A. -3 B. -1 C. 1 D. 3

We need to find the value of t that is less than -2. Let's go through each option:

  • A. -3: Is -3 less than -2? Yes, it is! So, this is a potential answer.
  • B. -1: Is -1 less than -2? No, it's greater than -2. So, this is not the correct answer.
  • C. 1: Is 1 less than -2? Absolutely not. 1 is a positive number, so it's much greater than -2. This is not the answer.
  • D. 3: Is 3 less than -2? Nope, 3 is also a positive number and much greater than -2. This is not the answer either.

From our evaluation, only option A, -3, satisfies the condition that t must be less than -2 for the equation to have no real solutions. Therefore, the correct answer is A. -3.

Conclusion: Mastering Quadratic Equations

Great job, everyone! We've successfully navigated through this quadratic equation problem. We started by understanding the basics of quadratic equations and the importance of the discriminant. Then, we applied the discriminant to our specific equation, 2x2−4x=t2x^2 - 4x = t, and determined that for the equation to have no real solutions, t must be less than -2. Finally, we evaluated the answer choices and found that -3 is the only value that satisfies this condition.

This problem highlights the power of the discriminant in determining the nature of the roots of a quadratic equation. By understanding and applying this concept, we can solve a wide range of problems involving quadratic equations. Remember, the key is to first rewrite the equation in the standard form (ax2+bx+c=0ax^2 + bx + c = 0), identify the coefficients a, b, and c, and then calculate the discriminant (b2−4acb^2 - 4ac). If the discriminant is less than 0, the equation has no real solutions; if it's equal to 0, there's one real solution; and if it's greater than 0, there are two real solutions.

Keep practicing with quadratic equations, and you'll become a pro in no time! Understanding these fundamental concepts is crucial for success in algebra and beyond. So, keep up the great work, and I'll see you in the next math adventure!