Question

Find all points on the x-axis that are 14 units from the point (4, -7).

(Simplify your answer. Type an ordered pair. Use a comma to separate answers as needed.)

Answer 1

The points are: (16.12,0),(-8.12,0).

Explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = a(x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\bigtriangleup))/(2*a)`

`\bigtriangleup = b^(2) - 4ac`

Distance between two points:

Suppose we have two points,
`(x_1,y_1)` and
`(x_2,y_2)`. The distance between them is given by:

`D = √((x_2-x_1)^2+(y_2-y_1)^2)`

Find all points on the x-axis that are 14 units from the point (4, -7).

Being on the x-axis mean that they have y-coordinate equal to 0, so the point is (x,0).

The distance is 14. So

`√((x-4)^2+(0-(-7))^2) = 14`

`√(x^2 - 8x + 16 + 49) = 14`

`√(x^2 - 8x + 65) = 14`

`(√(x^2 - 8x + 65))^2 = 14^2`

`x^2 - 8x + 65 - 196 = 0`

`x^2 - 8x - 131 = 0`

So
`a = 1, b = -8, c = -131`

`\bigtriangleup = (-8)^(2) - 4(1)(-131) = 588`

`x_(1) = (-(-8) + √(588))/(2) = 16.12`

`x_(1) = (-(-8) - √(588))/(2) = -8.12`

The points are: (16.12,0),(-8.12,0).