Find the x-intercepts?

Question

What are the x-intercepts for the equation y/9=250x cubed+150x squared+ 55x+70?

Answer 1

y/9 = 250x^3 + 150 x^2 + 55x + 70

To get the x-intercepts, we set y=0:

250x^3 + 150 x^2 + 55x + 70 = 0

50x^3 + 30x^2 + 11x + 14 = 0

Let f(x) = 50x^3 + 30x^2 + 11x + 14

We want to see where f(x)=0. We note that f(0) = 14, and that for all positive x, f(x) > 0, so there are no positive roots. Also, f(-1) = -50 + 30 - 11 + 14 = -17, so there's a root between -1 and 0. Also, for all x < -1, f(x) < 0, so there are no roots less than -1. All the roots are between -1 and 0.

We have two points, (0,14) and (-1,-17). Let's try x = -0.5:

f(-0.5) = 9.75 ==> (-0.5, 9.75) (I'm sort of interpolating)

f(-0.7) = 3.85 ==> (-0.7, 3.85)

f(-0.8) = -1.2 ==> (-0.8,-1.2)

f(-0.78) = -0.06 ==> (-0.78, -0.06)

That's pretty good, although we could go to a third decimal if we wanted to. Just as a check, f(-0.785) = -0.34. Let's go with one root at x = -0.78.

Now let's divide that root out, reducing the cubic to a quadratic:

f(x) = (x + 0.78)(50x^2 - 9x + 18) = 0

The roots of the quadratic g(x) = 50x^2 - 9x + 18 = 0 are imaginary. Using the quadratic formula,

x = 0.01 [9 +/- sqrt(81 - 3600)] = .01 [9 +/- 3 sqrt(9 - 400)]

x = .03 (3 +/- i sqrt 391)

so your only real root is x = -0.78. (Answer)