Question

The number of integral values of 'a' for which the quadratic equation \( ax^2 + ax + 5 = 0 \) cannot have real roots is

• A. Infinite
• B. 20
• C. 19
• D. 5

Hint:

Remember that for an equation to be "quadratic", the leading coefficient cannot be zero. Even if \( a=0 \) satisfied the inequality (it doesn't here, \( 0 \nless 0 \)), it would transform the equation into a linear one or a false statement (\( 5=0 \)).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

For a quadratic equation \( Ax^2 + Bx + C = 0 \) to have no real roots, its discriminant \( D \) must be strictly less than zero (\( D \textless 0 \)). Also, for it to be a quadratic equation, the coefficient of \( x^2 \) must not be zero (\( A \neq 0 \)).
Step 2: Key Formula or Approach:

Discriminant \( D = B^2 - 4AC \). Condition for no real roots: \( D \textless 0 \).
Step 3: Detailed Explanation:

Given equation: \( ax^2 + ax + 5 = 0 \). Here, \( A = a \), \( B = a \), \( C = 5 \). Calculate the discriminant: \[ D = (a)^2 - 4(a)(5) = a^2 - 20a \] We require \( D \textless 0 \): \[ a^2 - 20a \textless 0 \] \[ a(a - 20) \textless 0 \] The roots of \( a(a-20)=0 \) are \( a=0 \) and \( a=20 \). Since the quadratic in 'a' opens upwards, the expression is negative between the roots. \[ 0 \textless a \textless 20 \] The integral values of \( a \) in this interval are \( \{1, 2, 3, \dots, 19\} \). Note: We must check the condition \( A \neq 0 \). Since the interval is \( (0, 20) \), \( a=0 \) is already excluded. The number of such integer values is: \[ 19 - 1 + 1 = 19 \]
Step 4: Final Answer:

There are 19 integral values.