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Elementary College Geometry
Question 5 Page 7
View attachment 464
AB = BC
x^2 - 10 = 3x
x^2 - 3x -10 = 0
Solving for x, I got x = -2, x =5.
I know that AB is distance and so AB must be positive.
I will test each x-value to see which one yields a positive value.
AB = x^2 -10
Let x = -2
AB = (-2)^2 - 10
AB = 4 -10
AB = -6. . .reject
Let x = 5
AB = (5)^2 - 10
AB = 25 - 10
AB = 15
Question 5 Page 7
View attachment 464
AB = BC
x^2 - 10 = 3x
x^2 - 3x -10 = 0
Solving for x, I got x = -2, x =5.
I know that AB is distance and so AB must be positive.
I will test each x-value to see which one yields a positive value.
AB = x^2 -10
Let x = -2
AB = (-2)^2 - 10
AB = 4 -10
AB = -6. . .reject
Let x = 5
AB = (5)^2 - 10
AB = 25 - 10
AB = 15
