Source: 2018 Canadian Open Math Challenge Part A Problem 1 Suppose $x$ is a real number such that $x(x+3)=154.$ Determine the value of $(x+1)(x+2)$.
Problem
Source:
Tags: Comc, 2018 COMC
NikoIsLife
06.12.2018 17:30
$$x(x+3)=154\implies x^2+3x-154=0\implies(x+14)(x-11)=0\implies x=-14,11$$Both of these values give $(x+1)(x+2)=\boxed{156}$
\begin{align*}x(x+3)&=154\\x^2+3x&=154\\x^2+3x+2&=156\\(x+1)(x+2)&=\boxed{156}\end{align*}
firebolt360
06.12.2018 18:33
This is just a binomial!
$x(x+3) = 154$, so $x^2 + 3x -154 = 0$
Factor this to find the real value for x, and then plug it in to the other equation!
Or, you can see that $154$ is $14*11$ . $14 = 11 + 3$, so $x=11$. Remember, when you can't do something smart, do something stupid.
Either way, $12 \cdot 13 = 156$
Yajat
28.01.2024 09:25
x(x+3)=154, we know 154=11*14 and this matches the equation so x=11 and 12*13=156.
QueenArwen
13.08.2024 18:02
$x(x+3) = x^2+3x = 154.$ $(x+1)(x+2) = x^2+3x+2 = 154+2= 156$