Full solution

Q. Each of

a

and

b

can take values

1

2

with equal probabilty. The probability that the equation

ax^2+bx+1=0

has real roots, is equal to

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 ( $\Delta$ ) is given by $\Delta = b^2 - 4ac$ . Real roots exist if and only if the discriminant is greater than or equal to zero ( $\Delta \geq 0$ ).
Calculate Discriminant: Let's calculate the discriminant for the given equation: $\Delta = b^2 - 4\cdot a\cdot 1$ . 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: $\newline$ $1$ . For $(a, b) = (1, 1)$ : $\Delta = 1^2 - 4\cdot1\cdot1 = 1 - 4 = -3$ , which is less than $0$ . $\newline$ $2$ . For $(a, b) = (1, 2)$ : $\Delta = 2^2 - 4\cdot1\cdot1 = 4 - 4 = 0$ , which is equal to $0$ . $\newline$ $3$ . For $(a, b) = (2, 1)$ : $\Delta = 1^2 - 4\cdot2\cdot1 = 1 - 8 = -7$ , which is less than $0$ . $\newline$ $4$ . For $(a, b) = (2, 2)$ : $\Delta = 1^2 - 4\cdot1\cdot1 = 1 - 4 = -3$ $0$ , 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 $\frac{1}{4}$ . Therefore, the probability that the equation has real roots is the probability of the single combination that allows for real roots, which is $\frac{1}{4}$ .