The number of distinct quadratic equations ax | vedclass

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(B) For the quadratic equation $ax^2 + bx + c = 0$ to have unequal real roots,the discriminant $D = b^2 - 4ac$ must be greater than $0$ $(D > 0)$.Also

Vedclass · Accepted Answer

$6$

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(B) For the quadratic equation $ax^2 + bx + c = 0$ to have unequal real roots,the discriminant $D = b^2 - 4ac$ must be greater than $0$ $(D > 0)$.Also,for it to be a quadratic equation,$a 
eq 0$.Given the set $S = \{0, 1, 2, 4\}$ and the condition $a 
eq b 
eq c$,we test combinations of $(a, b, c)$ where $a \in \{1, 2, 4\}$ and $b, c \in \{0, 1, 2, 4\}$.$1$. If $a=1$: $b^2 - 4c > 0 \implies b^2 > 4c$. Possible $(b, c)$ pairs from $\{0, 2, 4\}$ (since $b, c 
eq 1$): - $(b=2, c=0) \implies 4 > 0$ (Valid)- $(b=4, c=0) \implies 16 > 0$ (Valid)- $(b=4, c=2) \implies 16 > 8$ (Valid)$2$. If $a=2$: $b^2 - 8c > 0 \implies b^2 > 8c$. Possible $(b, c)$ pairs from $\{0, 1, 4\}$ (since $b, c 
eq 2$): - $(b=1, c=0) \implies 1 > 0$ (Valid)- $(b=4, c=0) \implies 16 > 0$ (Valid)- $(b=4, c=1) \implies 16 > 8$ (Valid)$3$. If $a=4$: $b^2 - 16c > 0 \implies b^2 > 16c$. Possible $(b, c)$ pairs from $\{0, 1, 2\}$ (since $b, c 
eq 4$): - $(b=1, c=0) \implies 1 > 0$ (Valid)- $(b=2, c=0) \implies 4 > 0$ (Valid)Total valid equations = $3 + 3 + 2 = 8$. However,checking the constraint $a 
eq b 
eq c$ strictly,we re-evaluate: Valid sets $(a, b, c)$ are $(1, 2, 0), (1, 4, 0), (1, 4, 2), (2, 1, 0), (2, 4, 0), (2, 4, 1), (4, 1, 0), (4, 2, 0)$. All satisfy $a 
eq b, b 
eq c, a 
eq c$. Total is $8$. Since $8$ is not an option,re-checking the set: The question implies $a, b, c$ are distinct. The count is $8$.

The number of distinct quadratic equations $ax^2 + bx + c = 0$ with unequal real roots that can be formed by choosing the coefficients $a, b, c$ such that $a \neq b \neq c$ from the set $\{0, 1, 2, 4\}$ is:

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