Let m and n be two integers such that 0 leq m | vedclass

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(B) For the quadratic equation $x^2+mx+n=0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$. $D = m^2 - 4n \geq 0$,which

Vedclass · Accepted Answer

$73$

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(B) For the quadratic equation $x^2+mx+n=0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$. $D = m^2 - 4n \geq 0$,which implies $m^2 \geq 4n$. Given $0 \leq m, n \leq 10$,we count the pairs $(m, n)$ satisfying $n \leq \frac{m^2}{4}$: - If $m=10$,$n \leq 25$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values). - If $m=9$,$n \leq 20.25$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values). - If $m=8$,$n \leq 16$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values). - If $m=7$,$n \leq 12.25$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values). - If $m=6$,$n \leq 9$,so $n \in \{0, 1, \dots, 9\}$ ($10$ values). - If $m=5$,$n \leq 6.25$,so $n \in \{0, 1, \dots, 6\}$ ($7$ values). - If $m=4$,$n \leq 4$,so $n \in \{0, 1, 2, 3, 4\}$ ($5$ values). - If $m=3$,$n \leq 2.25$,so $n \in \{0, 1, 2\}$ ($3$ values). - If $m=2$,$n \leq 1$,so $n \in \{0, 1\}$ ($2$ values). - If $m=1$,$n \leq 0.25$,so $n \in \{0\}$ ($1$ value). - If $m=0$,$n \leq 0$,so $n \in \{0\}$ ($1$ value). Total number of pairs $= 11+11+11+11+10+7+5+3+2+1+1 = 73$.

Let $m$ and $n$ be two integers such that $0 \leq m \leq 10$ and $0 \leq n \leq 10$. Then,the number of ordered pairs $(m, n)$ such that $x^2+m x+n=0$ has real roots is

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