Question

Let t be 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

• A. $4t+3$
• B. $8t+5$
• C. $10t+3$
• D. $6t+5$

Answer 1

Answer: (B)

$8t+5$

Given for same positive integer $a$ and $b$

$t^1=at+b.....\left(1\right)$

Now,

since $t^3=t\times t^2$

$\Rightarrow t^3=t\times (at+b)$ (By using equation $\left(1\right)$)

$\Rightarrow t^3=a{t}^2+bt$

$\Rightarrow t^3=a(at+b)+bt$ (again by using equation $\left(1\right)$)

$\Rightarrow t^3=a^{2}t+ab+bt$

$\Rightarrow t^3=(a^2+b)t+ab$

By observing given, we can choose

$ab=3$ and $ab=5$

So, if $ab=3$ then there are two possible choice of $a$ and $b$

$\left(i\right)\frac{a=1}{t^3=(1^2+3)t+1\times 3}\left(ii\right)\frac{a=3,b=1}{t^3=(3^2+1)t+3}$$

$t^3=4+3t^3=10t+3$

$\therefore$ Option $A$ and $C$ are not correct.

Now, if $ab=5$ then there are two possible choices of $a$ & $b$

$\left(I\right)\frac{a=1b=5}{t^3=(1^2+5)t+5}\left(II\right)\frac{a=5,b=1}{t^3=(5^2+1)t+5}$

$t^3=6t+5t^3=26t+5$

So, clearly $t^3\ne 8t+5$

i.e. $t^3$ never equal to $8t+5$