If the coefficients a and b of a quadratic ex | vedclass

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(B) The quadratic equation $x^2+ax+b=0$ has real roots if the discriminant $D = a^2 - 4b \geq 0$, which implies $a^2 \geq 4b$.Given $a \in \{3, 4, 5\}

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$\frac{5}{6}$

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(B) The quadratic equation $x^2+ax+b=0$ has real roots if the discriminant $D = a^2 - 4b \geq 0$, which implies $a^2 \geq 4b$.Given $a \in \{3, 4, 5\}$ and $b \in \{1, 2, 3, 4\}$, the total number of possible pairs $(a, b)$ is $3 	imes 4 = 12$.We check the condition $a^2 \geq 4b$ for each pair:If $a=3$, $a^2=9$: $9 \geq 4b \implies b \leq 2.25$. Possible values for $b$ are $1, 2$ ($2$ pairs).If $a=4$, $a^2=16$: $16 \geq 4b \implies b \leq 4$. Possible values for $b$ are $1, 2, 3, 4$ ($4$ pairs).If $a=5$, $a^2=25$: $25 \geq 4b \implies b \leq 6.25$. Possible values for $b$ are $1, 2, 3, 4$ ($4$ pairs).Total favorable outcomes = $2 + 4 + 4 = 10$.Therefore, the required probability is $\frac{10}{12} = \frac{5}{6}$.

If the coefficients $a$ and $b$ of a quadratic expression $x^2+ax+b$ are chosen from the sets $A=\{3, 4, 5\}$ and $B=\{1, 2, 3, 4\}$ respectively, then the probability that the equation $x^2+ax+b=0$ has real roots is

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