+0  
 
0
8
1
avatar+952 

If $s$ is a real number, then what is the smallest possible value of $2s^2 - 8s + 19 - 7s^2 - 8s + 20$?

 Apr 9, 2024

Best Answer 

 #1
avatar+9664 
+1

I assume you meant largest.

Simplifying the expression and completing the square, we have

 

\(2s^2 - 8s + 19 - 7s^2 - 8s + 20\\ = -5s^2 - 16s + 39\\ = -5(s^2 + \frac{16}5s) + 39\\ = -5(s + \frac85)^2 + 5(\frac 85)^2 + 39\\ =-5(s + \frac85)^2 + \frac{259}5\)

 

Note that \(-5(s + \frac85)^2 \leq 0\) for any real s. Then the largest possible value is \(\dfrac{259}5\).

There is no smallest value. The expression can be arbitrarily small.

 Apr 9, 2024
 #1
avatar+9664 
+1
Best Answer

I assume you meant largest.

Simplifying the expression and completing the square, we have

 

\(2s^2 - 8s + 19 - 7s^2 - 8s + 20\\ = -5s^2 - 16s + 39\\ = -5(s^2 + \frac{16}5s) + 39\\ = -5(s + \frac85)^2 + 5(\frac 85)^2 + 39\\ =-5(s + \frac85)^2 + \frac{259}5\)

 

Note that \(-5(s + \frac85)^2 \leq 0\) for any real s. Then the largest possible value is \(\dfrac{259}5\).

There is no smallest value. The expression can be arbitrarily small.

MaxWong Apr 9, 2024

1 Online Users