X = -8 ± √(8)^2 - 4(1)(12) ---------- 2(1)

1 answer:

vladimir1956 [14]3 years ago

7 0

$x=\frac{-8\pm\sqrt{8^2-4(1)(12)}}{2(1)}\\x=\frac{-8\pm\sqrt{64-48}}{2}\\x=\frac{-8\pm\sqrt{16}}{2}\\x=\frac{-8\pm\sqrt{16}}{2}\\x =\frac{-8\pm4}{2}\\x=\frac{-8+4}{2} \ or \ \frac{-8-4}{2}\\x=\frac{-4}{2} \ or \ \frac{-12}{2}\\x=-2 \ or \ x=-6$

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Suppose that receiving stations X, Y, and Z are located on a coordinate plane at the points (3,3), (-13, −5), and (-5,3)

Aliun [14]

This problem will be solved both analytically and graphically.
The epicenter lies on a circle with radius=5 from X(3,3). Therefore
(x-3)² + (y-3)² = 25 (1)
Y(-13,-5) has the epicenter on a circle with radius=13, therefore
(x+13)² + (y+5)² = 169 (2)
Z (-5,3) has the epicenter on a circle with radius = 5, therefore
(x+5)² + (y-3)² = 25 (3)

Subtract (1) from (3).
(x+5)² - (x-3)² = 0
x² + 10x + 25 - x² + 6x - 9 = 0
16x = -16
x = -1
From (1), obtain
16 + (y-3)² = 25
(y-3)² = 9
y = 0 or y = 6

Check answers with (2).
When x=-1, y=0. obtain
(x+13)² + (y+5)² = 169 (Correct, Accept)
When x=-1, y=6, obtain
(x+13)² + (y+5)² = 265 (Incorrect, Reject)

The epicenter is at (-1,0).

Graphical solution (see the figure below).
From X (3,3), draw a circle with radius =5.
From Z (-5, 3), draw a circle with radius = 5.
The only point where all three circles intersect is (-1, 0) approximately.

Answer: The epicenter is at (-1,0)