Understanding the Nature of Solutions for a Quadratic Equation

Oct 13, 2025 by orovajewels.com 3 views

Introduction

In mathematics, a quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. The general form of a quadratic equation is $ax^2 + bx + c = 0$ , where $a$ , $b$ , and $c$ are constants. In this article, we will focus on determining the value of $K$ for a specific quadratic equation $4x^2 - 2x + K^2 - 2K + 1 = 0$ that has exactly one solution.

Understanding the Nature of Solutions for a Quadratic Equation

A quadratic equation can have either two distinct solutions, one repeated solution, or no real solutions. The nature of the solutions depends on the discriminant, which is the expression under the square root in the quadratic formula. The quadratic formula is given by:

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

For a quadratic equation to have exactly one solution, the discriminant must be equal to zero. This means that the expression under the square root, $b^2 - 4ac$ , must be equal to zero.

Applying the Condition for Exactly One Solution

For the given quadratic equation $4x^2 - 2x + K^2 - 2K + 1 = 0$ , we need to apply the condition for exactly one solution. The coefficients of the quadratic equation are $a = 4$ , $b = -2$ , and $c = K^2 - 2K + 1$ . We need to set the discriminant equal to zero and solve for $K$ .

Setting Up the Equation for the Discriminant

The discriminant is given by the expression $b^2 - 4ac$ . Substituting the values of $a$ , $b$ , and $c$ , we get:

(-2)^2 - 4(4)(K^2 - 2K + 1) = 0

Simplifying the expression, we get:

4 - 16(K^2 - 2K + 1) = 0

Solving for $K$

Expanding the expression, we get:

4 - 16K^2 + 32K - 16 = 0

Combining like terms, we get:

-16K^2 + 32K - 12 = 0

Dividing the entire equation by -4, we get:

4K^2 - 8K + 3 = 0

Using the Quadratic Formula to Solve for $K$

We can use the quadratic formula to solve for $K$ . The quadratic formula is given by:

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

Substituting the values of $a$ , $b$ , and $c$ , we get:

K = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(4)(3)}}{2(4)}

Simplifying the expression, we get:

K = \frac{8 \pm \sqrt{64 - 48}}{8}

K = \frac{8 \pm \sqrt{16}}{8}

K = \frac{8 \pm 4}{8}

Finding the Values of $K$

We have two possible values for $K$ :

K = \frac{8 + 4}{8} = \frac{12}{8} = \frac{3}{2}

K = \frac{8 - 4}{8} = \frac{4}{8} = \frac{1}{2}

Conclusion

In this article, we determined the value of $K$ for a quadratic equation $4x^2 - 2x + K^2 - 2K + 1 = 0$ that has exactly one solution. We applied the condition for exactly one solution, set up the equation for the discriminant, and solved for $K$ using the quadratic formula. We found two possible values for $K$ : $\frac{3}{2}$ and $\frac{1}{2}$ . These values of $K$ ensure that the quadratic equation has exactly one solution.

Future Work

In future work, we can explore other quadratic equations and determine the values of their coefficients that result in exactly one solution. We can also investigate the properties of the solutions and how they relate to the coefficients of the quadratic equation.

References

[1] "Quadratic Equation" by Math Open Reference. Retrieved from https://www.mathopenref.com/quadratic.html
[2] "Discriminant" by Wolfram MathWorld. Retrieved from https://mathworld.wolfram.com/Discriminant.html
[3] "Quadratic Formula" by Khan Academy. Retrieved from https://www.khanacademy.org/math/algebra/x2f-quadratic-formula/x2f-quadratic-formula-article/v/quadratic-formula-article

Introduction

In our previous article, we determined the value of $K$ for a quadratic equation $4x^2 - 2x + K^2 - 2K + 1 = 0$ that has exactly one solution. We applied the condition for exactly one solution, set up the equation for the discriminant, and solved for $K$ using the quadratic formula. In this article, we will answer some frequently asked questions (FAQs) related to quadratic equations with exactly one solution.

Q: What is the condition for a quadratic equation to have exactly one solution?

A: The condition for a quadratic equation to have exactly one solution is that the discriminant must be equal to zero. This means that the expression under the square root in the quadratic formula must be equal to zero.

Q: How do I determine the value of $K$ for a quadratic equation with exactly one solution?

A: To determine the value of $K$ for a quadratic equation with exactly one solution, you need to apply the condition for exactly one solution, set up the equation for the discriminant, and solve for $K$ using the quadratic formula.

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that provides the solutions to a quadratic equation. It is given by:

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

Q: How do I use the quadratic formula to solve for $K$ ?

A: To use the quadratic formula to solve for $K$ , you need to substitute the values of $a$ , $b$ , and $c$ into the formula and simplify the expression.

Q: What are the possible values of $K$ for a quadratic equation with exactly one solution?

A: The possible values of $K$ for a quadratic equation with exactly one solution depend on the specific equation. In our previous article, we found two possible values of $K$ : $\frac{3}{2}$ and $\frac{1}{2}$ .

Q: Can a quadratic equation have more than one solution?

A: Yes, a quadratic equation can have more than one solution. In fact, a quadratic equation can have two distinct solutions, one repeated solution, or no real solutions.

Q: How do I determine the nature of the solutions for a quadratic equation?

A: To determine the nature of the solutions for a quadratic equation, you need to examine the discriminant. If the discriminant is positive, the equation has two distinct solutions. If the discriminant is zero, the equation has one repeated solution. If the discriminant is negative, the equation has no real solutions.

Q: Can a quadratic equation have no real solutions?

A: Yes, a quadratic equation can have no real solutions. This occurs when the discriminant is negative.

Q: How do I determine the value of $K$ for a quadratic equation with no real solutions?

A: To determine the value of $K$ for a quadratic equation with no real solutions, you need to examine the discriminant. If the discriminant is negative, you can use the quadratic formula to find the complex solutions.

Conclusion

In this article, we answered some frequently asked questions (FAQs) related to quadratic equations with exactly one solution. We discussed the condition for a quadratic equation to have exactly one solution, how to determine the value of $K$ for a quadratic equation with exactly one solution, and how to use the quadratic formula to solve for $K$ . We also discussed the nature of the solutions for a quadratic equation and how to determine the value of $K$ for a quadratic equation with no real solutions.

Future Work

References

[1] "Quadratic Equation" by Math Open Reference. Retrieved from https://www.mathopenref.com/quadratic.html
[2] "Discriminant" by Wolfram MathWorld. Retrieved from https://mathworld.wolfram.com/Discriminant.html
[3] "Quadratic Formula" by Khan Academy. Retrieved from https://www.khanacademy.org/math/algebra/x2f-quadratic-formula/x2f-quadratic-formula-article/v/quadratic-formula-article