The Nature of Roots of a Quadratic Equation

Oct 13, 2025 by sorusgentrification.com 113 views

Introduction

In mathematics, the nature of roots of a quadratic equation plays a crucial role in understanding the behavior of the equation. A quadratic equation is of the form $ax^2 + bx + c = 0$ , where $a$ , $b$ , and $c$ are constants. The nature of roots of a quadratic equation depends on the discriminant, which is given by the expression $b^2 - 4ac$ . In this article, we will discuss the application of your knowledge on the nature of roots of a quadratic equation, specifically the equation $x^2 + (1-k)x + k-3 = 0$ . We will prove that the roots of this equation are real for all real values of $k$ .

The Nature of Roots of a Quadratic Equation

The nature of roots of a quadratic equation can be determined using the discriminant. If the discriminant is positive, then the roots are real and distinct. If the discriminant is zero, then the roots are real and equal. If the discriminant is negative, then the roots are complex and distinct.

The Quadratic Equation $x^2 + (1-k)x + k-3 = 0$

The given quadratic equation is $x^2 + (1-k)x + k-3 = 0$ . To determine the nature of roots of this equation, we need to calculate the discriminant.

Calculating the Discriminant

The discriminant of the quadratic equation $x^2 + (1-k)x + k-3 = 0$ is given by the expression $(1-k)^2 - 4(1)(k-3)$ . We can simplify this expression as follows:

$(1-k)^2 - 4(1)(k-3)$ $= 1 - 2k + k^2 - 4k + 12$ $= k^2 - 6k + 13$

Analyzing the Discriminant

Now that we have calculated the discriminant, we need to analyze it to determine the nature of roots of the quadratic equation. The discriminant is a quadratic expression in $k$ , and we can analyze it by considering its sign.

Case 1: $k^2 - 6k + 13 \geq 0$

If $k^2 - 6k + 13 \geq 0$ , then the discriminant is non-negative, and the roots of the quadratic equation are real and distinct.

Case 2: $k^2 - 6k + 13 < 0$

If $k^2 - 6k + 13 < 0$ , then the discriminant is negative, and the roots of the quadratic equation are complex and distinct.

Case 3: $k^2 - 6k + 13 = 0$

If $k^2 - 6k + 13 = 0$ , then the discriminant is zero, and the roots of the quadratic equation are real and equal.

Conclusion

In conclusion, we have analyzed the discriminant of the quadratic equation $x^2 + (1-k)x + k-3 = 0$ and determined that the roots of this equation are real for all real values of $k$ . This is because the discriminant is always non-negative, regardless of the value of $k$ . Therefore, the roots of the quadratic equation are always real and distinct.

Proof

To prove that the roots of the quadratic equation $x^2 + (1-k)x + k-3 = 0$ are real for all real values of $k$ , we need to show that the discriminant is always non-negative.

Let $f(k) = k^2 - 6k + 13$ . We can show that $f(k) \geq 0$ for all real values of $k$ by considering the following cases:

If $k \leq 3$ , then $f(k) = k^2 - 6k + 13 \geq 0$ .
If $k > 3$ , then $f(k) = k^2 - 6k + 13 = (k-3)^2 + 4 \geq 0$ .

Therefore, we have shown that $f(k) \geq 0$ for all real values of $k$ , and the discriminant is always non-negative. This implies that the roots of the quadratic equation $x^2 + (1-k)x + k-3 = 0$ are real for all real values of $k$ .

Final Conclusion

In conclusion, we have analyzed the nature of roots of the quadratic equation $x^2 + (1-k)x + k-3 = 0$ and proved that the roots of this equation are real for all real values of $k$ . This is a significant result, as it implies that the quadratic equation has real roots for all possible values of $k$ .