Question

Let $f\left(x\right)=x^2+x-6$. For what values of "$t$" does $f(t-5)=0$?

• A. $-3$ and $2$
• B. $-2$ and $3$
• C. $5$
• D. $2$ and $7$

Answer 1

Answer: (D)

$2$ and $7$

$f\left(x\right)=x^2+x-6$

$\therefore f\left(t\right)=t^2+t-6$

$\therefore f(t-5)=(t-5{)}^2+(t-5)-6$

$\therefore f(t-5)=t^2-9t+14$

$f(t-5)=0$

$⟹t^2-9t+14=0$

$\therefore t^2-7t-2t-14=0$

$\therefore t(t-7)-2(t-7)=0$

$\therefore (t-7)(t-2)=0$

$\therefore t=7$ or $t=2$

So, the answer is option (D)