Math Problem: Finding The Least Integer For No Real Solutions

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Hey guys! Let's dive into a cool math problem that's perfect for flexing those brain muscles. We're going to break down a quadratic equation and figure out when it doesn't have any real solutions. This is super important stuff for anyone prepping for standardized tests like the SAT or ACT, or even just wanting to keep their math skills sharp. So, grab your pencils, and let's get started!

Understanding the Core Concept: Quadratic Equations and Real Solutions

First things first, let's make sure we're all on the same page about quadratic equations. A quadratic equation is a fancy way of saying an equation that can be written in the form ax2+bx+c=0{ax^2 + bx + c = 0}, where a{a}, b{b}, and c{c} are constants, and a{a} is not equal to zero. These equations are super common in math and have a cool graphical representation: they form parabolas (U-shaped curves). The solutions to a quadratic equation are the values of x{x} that make the equation true. These are also known as the roots or zeros of the equation, and they represent where the parabola crosses the x-axis.

Now, here's where it gets interesting. A quadratic equation can have:

  • Two distinct real solutions: The parabola crosses the x-axis at two different points.
  • One real solution (a repeated root): The parabola touches the x-axis at exactly one point (the vertex of the parabola is on the x-axis).
  • No real solutions: The parabola doesn't cross the x-axis at all (it hovers either entirely above or entirely below the x-axis).

The key to figuring out the number of real solutions lies in something called the discriminant. The discriminant is the part of the quadratic formula under the square root: b2−4ac{b^2 - 4ac}. The discriminant tells us the nature of the roots:

  • If b2−4ac>0{b^2 - 4ac > 0}, there are two distinct real roots.
  • If b2−4ac=0{b^2 - 4ac = 0}, there is one real root (a repeated root).
  • If b2−4ac<0{b^2 - 4ac < 0}, there are no real roots.

So, when we're asked to find the conditions under which a quadratic equation has no real solutions, we're basically looking for when the discriminant is less than zero. That's our golden ticket!

To summarize: a quadratic equation has no real solutions when its discriminant is negative. This means b2−4ac<0{b^2 - 4ac < 0}.

Decoding the Problem: Setting Up the Equation

Alright, let's get down to brass tacks with the actual problem. We're given the equation: x(kx−68)=−4{x(kx - 68) = -4}. Our mission is to figure out the least possible integer value of k{k} such that this equation has no real solutions. The first thing we need to do is rearrange this equation into the standard quadratic form, ax2+bx+c=0{ax^2 + bx + c = 0}. This is crucial because it allows us to identify the coefficients a{a}, b{b}, and c{c}, which we'll need to calculate the discriminant.

Here’s how we do it step-by-step:

  1. Expand the equation: Multiply out the x{x} on the left side: kx2−68x=−4{kx^2 - 68x = -4}.
  2. Move everything to one side: Add 4 to both sides to get everything on the left side and set the equation equal to zero: kx2−68x+4=0{kx^2 - 68x + 4 = 0}.

Now we have the equation in the standard quadratic form: kx2−68x+4=0{kx^2 - 68x + 4 = 0}. We can easily identify the coefficients:

  • a=k{a = k}
  • b=−68{b = -68}
  • c=4{c = 4}

Now that the equation is in standard form, and we've identified the coefficients, we can move on to the next step, which involves using the discriminant to figure out the value of k{k} for which the equation has no real solutions. This is where the real fun begins, so stay with me!

Remember, we want the discriminant to be less than zero. Let's make this happen!

Solving for k: Using the Discriminant

Alright, folks, we're in the home stretch! We've got our quadratic equation in the right form, and we know our goal: to find the least integer value of k{k} that makes the discriminant negative (less than zero). Let's put our knowledge of the discriminant to work. Recall that the discriminant is b2−4ac{b^2 - 4ac}. Since we want no real solutions, we want:

b2−4ac<0{b^2 - 4ac < 0}

We know that a=k{a = k}, b=−68{b = -68}, and c=4{c = 4}. Substitute these values into the discriminant inequality:

(−68)2−4(k)(4)<0{(-68)^2 - 4(k)(4) < 0}

Now, let's simplify this inequality and solve for k{k}:

  1. Calculate (−68)2{(-68)^2}: (−68)2=4624{(-68)^2 = 4624}
  2. Substitute and simplify: 4624−16k<0{4624 - 16k < 0}
  3. Isolate k{k}: Subtract 4624 from both sides: −16k<−4624{-16k < -4624}
  4. Divide by −16{-16} (and remember to flip the inequality sign because we're dividing by a negative number): k>−4624−16{k > \frac{-4624}{-16}}
  5. Calculate the value: k>289{k > 289}

So, we find that k>289{k > 289}. This means that for the equation to have no real solutions, k{k} must be greater than 289. Since the question asks for the least possible integer value of k{k}, we want the smallest integer that is greater than 289. That integer is 290.

Therefore, the answer is C) 290!

Conclusion: The Final Answer and Key Takeaways

Awesome work, everyone! We've successfully navigated the math problem and figured out the least possible value of k{k} for which the given quadratic equation has no real solutions. To recap:

  • We understood the concept of quadratic equations and the importance of the discriminant in determining the nature of the roots.
  • We converted the given equation into standard quadratic form.
  • We applied the discriminant formula b2−4ac{b^2 - 4ac} and set it less than zero to find the range of values for k{k} that would yield no real solutions.
  • We found that the least integer value of k{k} that satisfies the condition is 290.

Key Takeaways:

  • Always rearrange the equation into standard quadratic form ax2+bx+c=0{ax^2 + bx + c = 0} to identify the coefficients.
  • Remember the discriminant's role: b2−4ac>0{b^2 - 4ac > 0} for two real roots, b2−4ac=0{b^2 - 4ac = 0} for one real root, and b2−4ac<0{b^2 - 4ac < 0} for no real roots.
  • Pay close attention to inequalities and remember to flip the inequality sign when multiplying or dividing by a negative number.

This problem is a great example of how understanding fundamental mathematical concepts can help you solve more complex problems. Keep practicing, and you'll be acing these types of questions in no time! Keep learning, keep growing, and keep conquering those math problems!

Now you're equipped to tackle similar problems with confidence. Keep up the awesome work!