Respuesta :
Answer:
The points are: (16.12,0),(-8.12,0).
Step-by-step explanation:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
Distance between two points:
Suppose we have two points, [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]. The distance between them is given by:
[tex]D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
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
[tex]\sqrt{(x-4)^2+(0-(-7))^2} = 14[/tex]
[tex]\sqrt{x^2 - 8x + 16 + 49} = 14[/tex]
[tex]\sqrt{x^2 - 8x + 65} = 14[/tex]
[tex](\sqrt{x^2 - 8x + 65})^2 = 14^2[/tex]
[tex]x^2 - 8x + 65 - 196 = 0[/tex]
[tex]x^2 - 8x - 131 = 0[/tex]
So [tex]a = 1, b = -8, c = -131[/tex]
[tex]\bigtriangleup = (-8)^{2} - 4(1)(-131) = 588[/tex]
[tex]x_{1} = \frac{-(-8) + \sqrt{588}}{2} = 16.12[/tex]
[tex]x_{1} = \frac{-(-8) - \sqrt{588}}{2} = -8.12[/tex]
The points are: (16.12,0),(-8.12,0).
