Algebra

Question

0

48

1

+1855

What real value of $t$ produces the smallest value of the quadratic $t^2 -9t - 36 + 13t - 60$?

ABJeIIy Apr 9, 2024

+9675

+1

Best Answer

Simplifying and completing the square, we have

$t^2 -9t - 36 + 13t - 60\\ = t^2 + 4t - 96\\ = t^2 + 4t + 4 - 100\\ = (t + 2)^2 - 100$

Note that $(t + 2)^2 \geq 0$ for any real t, and equality holds when $t + 2 = 0$ , i.e., $t = -2$ .

Hence, the smallest value is produced when t = -2.

MaxWong Apr 9, 2024

MaxWong · Answer 1 · 2024-04-09T03:53:15

Simplifying and completing the square, we have

$t^2 -9t - 36 + 13t - 60\\ = t^2 + 4t - 96\\ = t^2 + 4t + 4 - 100\\ = (t + 2)^2 - 100$

Note that $(t + 2)^2 \geq 0$ for any real t, and equality holds when $t + 2 = 0$ , i.e., $t = -2$ .

Hence, the smallest value is produced when t = -2.