Question

Let \(a\) be an integer selected at random from the set \(\{0,1,2,3,\ldots ,9\}\). The probability that the equation \(ax^2-ax+1 = 0\) has real roots is ...

• A. \(\frac{3}{5}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{2}{5}\)
• D. \(\frac{5}{9}\)

Hint:

A quadratic needs a nonzero a and discriminant at least zero; count the integers from 0 to 9 that qualify.

Answer 1

The Correct Option is A

Step 1: Plan:
Find which digits $a$ make the parabola $ax^2-ax+1$ touch or cross the x-axis.

Step 2: Work:
Discriminant $=a^2-4a$. It is zero at $a=0$ and $a=4$, negative for $0<a<4$, positive for $a>4$.
The case $a=0$ is excluded because the equation degenerates to $1=0$.

Step 3: Count:
Valid digits: 4, 5, 6, 7, 8, 9, which is 6 of 10, giving $\tfrac{6}{10}=\tfrac35$. Option (A).

Final Answer:
Six of the ten digits work, so the answer is (A). \[ \boxed{\text{(A) } \frac{3}{5}} \]