The expression $ax^2 + bx + 1$ takes the values $1$ and $4$ when $x$ takes the values $2$ and $3$ respectively. How can we find the value of the expression when $x$ takes the value of $4$?
$\endgroup$3 Answers
$\begingroup$Substitution shows that $4a+2b+1=1$ and $9a+3b+1=4$. Now solve the simultaneous equations to get the value of $a$ and $b$.
$\endgroup$ 0 $\begingroup$Let $f(x) = ax^2 + bx + 1$
Plug the values in $f(x)$ and we have:
$$f(2) = a(2)^2 + 2b + 1 = 4a + 2b + 1 = 1$$ $$f(3) = a(3)^2 + 3b + 1 = 9a + 3b + 1 = 4$$
Then just solve the following system of equations to find the values for $a$ and $b$, which is fairly easy:
\begin{cases} 4a + 2b = 0\\ 9a + 3b = 3 \end{cases}
Then just plug $x=4$ in $f(x)$.
$\endgroup$ $\begingroup$Hint $\,\ f(3)=2^2,\ f(2) = 1^2,\ f(0) = (-1)^2\ $ so $\,f-(x-1)^2\!=0,\,$ having $3$ roots $\,x=3,2,0$.
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