x2%2B2(k-12)x%2B2%3D0-find-k.jpeg "(k-12)x2+2(k-12)x+2=0 Find K")
Solving for 'k' in the Quadratic Equation (k-12)x² + 2(k-12)x + 2 = 0
This article will guide you through the process of solving for the value of 'k' in the given quadratic equation: (k-12)x² + 2(k-12)x + 2 = 0.
Understanding the Problem
The equation provided is a quadratic equation, meaning it has the general form ax² + bx + c = 0. In this case:
- a = (k-12)
- b = 2(k-12)
- c = 2
To find the value of 'k', we need to consider the nature of the roots of this quadratic equation.
Discriminant and Nature of Roots
The discriminant of a quadratic equation (Δ) is given by Δ = b² - 4ac. It tells us about the nature of the roots:
- Δ > 0: Two distinct real roots
- Δ = 0: One real root (repeated root)
- Δ < 0: Two complex roots
Since we are interested in finding the value of 'k', we can use the discriminant to find the conditions that will give us real roots (either distinct or repeated).
Solving for 'k'
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Calculate the discriminant (Δ): Δ = [2(k-12)]² - 4(k-12)(2) Δ = 4(k-12)² - 8(k-12) Δ = 4(k-12)(k-12-2) Δ = 4(k-12)(k-14)
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Set the discriminant greater than or equal to zero: We want real roots, so Δ ≥ 0. 4(k-12)(k-14) ≥ 0
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Solve the inequality: The inequality holds true when:
- k ≤ 12 or k ≥ 14
Therefore, the values of 'k' that will give the quadratic equation real roots are k ≤ 12 or k ≥ 14.
Conclusion
By using the discriminant and analyzing the nature of the roots, we have successfully found the values of 'k' that will give the quadratic equation real roots. Remember that these values will lead to either two distinct real roots or one real root (repeated root).