Question

Which equation is equivalent to $1-\frac{6}{t^2}=\frac{1}{t}$ ?

• A. $(t-3)(t+2)=0$
• B. $(t-2)(t+3)=0$
• C. $(2t+1)(3t-1)=0$
• D. $(2t-1)(3t+1)=0$

Answer 1

The correct option is A $(t-3)(t+2)=0$

Given, $1-\frac{6}{t^2}=\frac{1}{t}$
Multiplying both sides by $t^2$ where $t\ne 0$, we get

$t^2-6=t$
$\Rightarrow t^2-t-6=0$

$\Rightarrow t^2-3t+2t-6=0$
$\Rightarrow t(t-3)+2(t-3)=0$

$\Rightarrow (t-3)(t+2)=0$
Hence, option (a) is the correct choice.