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(A) For the quadratic equation $ax^2 - ax + 1 = 0$ to have real roots, the discriminant $D$ must be greater than or equal to $0$.$D = b^2 - 4ac = (-a)

Vedclass · Accepted Answer

$3/5$

Vedclass · Answer

(A) For the quadratic equation $ax^2 - ax + 1 = 0$ to have real roots, the discriminant $D$ must be greater than or equal to $0$.$D = b^2 - 4ac = (-a)^2 - 4(a)(1) = a^2 - 4a$.Condition: $a^2 - 4a \ge 0 \implies a(a - 4) \ge 0$.This inequality holds when $a \le 0$ or $a \ge 4$.Given the set $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, the total number of elements is $10$.If $a = 0$, the equation becomes $1 = 0$, which is false, so $a = 0$ does not yield real roots.For $a \ge 4$, the values are $\{4, 5, 6, 7, 8, 9\}$, which are $6$ values.Thus, the number of favorable outcomes is $6$.Probability $P = \frac{6}{10} = \frac{3}{5}$.

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

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