Let t be a real number such that t^2 = at + b | vedclass

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(B) Given $t^2 = at + b$,where $a$ and $b$ are positive integers.Multiplying by $t$,we get $t^3 = at^2 + bt$.Substituting $t^2 = at + b$ into the equa

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$8t + 5$

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(B) Given $t^2 = at + b$,where $a$ and $b$ are positive integers.Multiplying by $t$,we get $t^3 = at^2 + bt$.Substituting $t^2 = at + b$ into the equation:$t^3 = a(at + b) + bt = a^2t + ab + bt = (a^2 + b)t + ab$.We check each option for the form $(a^2 + b)t + ab$:$(A)$ $4t + 3$: $a^2 + b = 4$ and $ab = 3$. If $a = 1$,then $b = 3$,which satisfies $1^2 + 3 = 4$. Possible.$(B)$ $8t + 5$: $a^2 + b = 8$ and $ab = 5$. If $a = 1$,$b = 5$,then $a^2 + b = 6 
eq 8$. If $a = 5$,$b = 1$,then $a^2 + b = 26 
eq 8$. No positive integer solution exists. Not possible.$(C)$ $10t + 3$: $a^2 + b = 10$ and $ab = 3$. If $a = 3$,$b = 1$,then $a^2 + b = 9 + 1 = 10$. Possible.$(D)$ $6t + 5$: $a^2 + b = 6$ and $ab = 5$. If $a = 1$,$b = 5$,then $a^2 + b = 1 + 5 = 6$. Possible.Thus,$t^3$ is never equal to $8t + 5$.

Let $t$ be a real number such that $t^2 = at + b$ for some positive integers $a$ and $b$. Then,for any choice of positive integers $a$ and $b$,$t^3$ is never equal to:

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