Each of a and b can take values 1 or 2 with equal probabilty. The probability that the equation ax^2+bx+1=0 has real roots, is equal to

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Determine Real Roots Condition: We need to determine the conditions under which the quadratic equation ax^2 + bx + 1 = 0 has real roots. For a quadratic equation ax^2 + bx + c = 0, the discriminant (Δ) is given by Δ = b^2 - 4ac. Real roots exist if and only if the discriminant is greater than or equal to zero (Δ ≥ 0).Calculate Discriminant: Let's calculate the discriminant for the given equation: Δ = b^2 - 4*a*1. Since a and b can each be 1 or 2, we have four possible combinations for (a, b): (1, 1), (1, 2), (2, 1), and (2, 2).Check Combinations: For each combination, we will calculate the discriminant and check if it is non-negative: 1. For (a, b) = (1, 1): Δ = 1^2 - 4*1*1 = 1 - 4 = -3, which is less than 0. 2. For (a, b) = (1, 2): Δ = 2^2 - 4*1*1 = 4 - 4 = 0, which is equal to 0. 3. For (a, b) = (2, 1): Δ = 1^2 - 4*2*1 = 1 - 8 = -7, which is less than 0. 4. For (a, b) = (2, 2): Δ = 2^2 - 4*2*1 = 4 - 8 = -4, which is less than 0.Identify Real Roots: Out of the four combinations, only one combination (1, 2) results in a non-negative discriminant, which means the equation has real roots in this case.Calculate Probability: Since each of a and b can take values 1 or 2 with equal probability, each combination has an equal probability of 1/4. Therefore, the probability that the equation has real roots is the probability of the single combination that allows for real roots, which is 1/4.

Each of $a$ and $b$ can take values $1$ or $2$ with equal probabilty. The probability that the equation $ax^2+bx+1=0$ has real roots, is equal to

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Each of $a$ and $b$ can take values $1$ or $2$ with equal probabilty. The probability that the equation $ax^2+bx+1=0$ has real roots, is equal to