Factor each A.) x^3+7x^2+10x=0 B.) x^2 -6x+8=0
1 answer:
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Answer:
(-2, -3)
Step-by-step explanation:
A careful graph shows the point (-2, -3) is at the intersection of the circles whose radii are the given distances from the receiving stations.
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The simultaneous equations for the circles can be solved algebraically.
The epicenter is 10 units from X, so lies on the circle ...
(x -4)^2 +(y -5)^2 = 10^2
x^2 -8x +16 +y^2 -10y +25 = 100
x^2 +y^2 -8x -10y = 59
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The epicenter is 5 units from Y, so lies on the circle ...
(x +6)^2 +(y +6)^2 = 5^2
x^2 +12x +36 +y^2 +12y +36 = 25
x^2 +y^2 +12x +12y = -47
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The epicenter is 13 units from Z, so lies on the circle ...
(x +14)^2 +(y -2)^2 = 13^2
x^2 +28x +196 +y^2 -4y +4 = 169
x^2 +y^2 +28x -4y = -31
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Subtracting the second equation from each of the other two, we get ...
(x^2 +y^2 -8x -10y) -(x^2 +y^2 +12x +12y) = (59) -(-47)
-20x -22y = 106 . . . . eq1 -eq2
(x^2 +y^2 +28x -4y) -(x^2 +y^2 +12x +12y) = (-31) -(-47)
16x -16y = 16 . . . . . . . .eq3 -eq2
These simultaneous linear equations can be solved a variety of ways. We might use substitution:
x = y+1 . . . . . from eq3 -eq2 divided by 16
10(y +1) +11y = -53 . . . . . from eq1 -eq2 divided by -2
21y = -63 . . . . . . . . . . . . simplify, subtract 10
y = -3
x = y+1 = -2
The epicenter is located at (x, y) = (-2, -3).
