Algebra

Question

0

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+1855

If $t$ is a real number, what is the maximum possible value of the expression $-t^2 + 8t -4 +5t^2 - 4t + 18$?

ABJeIIy Apr 9, 2024

+9675

+1

Best Answer

I assume you mean minimum.

Simplifying and completing the square,

$-t^2 +8t-4+5t^2-4t+18\\ =4t^2 +4t + 14\\ =4t^2 +4t+1 +13\\ =(2t+1)^2 + 13\\$

Note that $(2t + 1)^2 \geq 0$ for all real t. Then the minimum is 13.

The expression can be arbitrarily large so there is no maximum.

MaxWong Apr 9, 2024

MaxWong · Answer 1 · 2024-04-09T03:48:57

I assume you mean minimum.

Simplifying and completing the square,

$-t^2 +8t-4+5t^2-4t+18\\ =4t^2 +4t + 14\\ =4t^2 +4t+1 +13\\ =(2t+1)^2 + 13\\$

Note that $(2t + 1)^2 \geq 0$ for all real t. Then the minimum is 13.

The expression can be arbitrarily large so there is no maximum.