Question

If the roots of the quadratic equation \(5x^2 - 10x + k = 0\) are real and equal, then the value of k is :

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

For a quadratic equation \(ax^2 + bx + c = 0\) to have equal roots, it must be a perfect square trinomial.
Rewriting: \(5(x^2 - 2x) + k = 0 \implies 5(x^2 - 2x + 1) = 0 \implies 5x^2 - 10x + 5 = 0\).
By comparing this directly with the given equation, we immediately see that \(k = 5\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question is from the topic of Quadratic Equations, specifically concerning the nature of roots.
We are given a quadratic equation with an unknown constant \(k\). We need to find \(k\) such that the equation has real and equal roots.

Step 2: Key Formula or Approach: For a standard quadratic equation \(ax^2 + bx + c = 0\), the nature of its roots depends on the value of the discriminant (\(D\)):
\[ D = b^2 - 4ac \]
The condition for the roots to be real and equal is:
\[ D = 0 \implies b^2 - 4ac = 0 \]

Step 3: Detailed Explanation:
Given quadratic equation:
\[ 5x^2 - 10x + k = 0 \]
Identify the coefficients:
\[ a = 5, \quad b = -10, \quad c = k \]
Apply the condition for real and equal roots:
\[ b^2 - 4ac = 0 \]
Substitute the values of \(a\), \(b\), and \(c\):
\[ (-10)^2 - 4(5)(k) = 0 \]
Calculate the terms:
\[ 100 - 20k = 0 \]
Rearrange the equation to solve for \(k\):
\[ 20k = 100 \]
\[ k = \frac{100}{20} \]
\[ k = 5 \]

Step 4: Final Answer:
The value of \(k\) is \(5\).