Question

For what value of k, roots of the quadratic equation \(kx^2 - 6x + 1 = 0\) are real and equal?

• A. 6
• B. 8
• C. 9
• D. 10

Hint:

Remember the conditions for the nature of roots based on the discriminant: - \(D > 0\): real and distinct roots - \(D = 0\): real and equal roots - \(D < 0\): no real roots This question specifically asks for the "real and equal" case, so \(D=0\) is the key.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The roots of a quadratic equation \(ax^2 + bx + c = 0\) are real and equal if and only if its discriminant, \(D\), is equal to zero.

Step 2: Key Formula or Approach:
The discriminant is given by \(D = b^2 - 4ac\). We need to set \(D = 0\) and solve for \(k\).

Step 3: Detailed Explanation:
For the given equation \(kx^2 - 6x + 1 = 0\), we have:
\(a = k\), \(b = -6\), \(c = 1\).
Set the discriminant to zero for real and equal roots:
\[ D = b^2 - 4ac = 0 \] \[ (-6)^2 - 4(k)(1) = 0 \] \[ 36 - 4k = 0 \] Now, solve for \(k\):
\[ 36 = 4k \] \[ k = \frac{36}{4} \] \[ k = 9 \]

Step 4: Final Answer:
The value of k for which the roots are real and equal is 9.