+1348 help im stuck
Chords $\overline{PQ}$ and $\overline{RS}$ of a circle meet at $X$ inside the circle. If $RS = 36$, $PX = 8$, and $QX = 24$, then what is the smallest possible value of $RX$?
+129771 R
m
P 8 X 24 Q
36 - m
S
Intersecting Chords Theorem
Let RX = m and SX = 36 - m
PX * QX = RX * SX
8 * 24 = m ( 36 - m)
192 = 36m - m^2
m^2 - 36m + 192 = 0
m^2 - 36m = -192 complete the square on m
m^2 - 36m + 324 = -192 + 324
(m - 18)^2 = 132 take the positive root
m - 18 = sqrt [ 132]
m = sqrt [132] + 18 = 2sqrt33 + 18 = 2 [ sqrt 33 + 9 ] = smallest positive value of RX
