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BabaBlast [244]
3 years ago
9

Three roots of the polynomial equation x4 - 4x3 - 2x2 + 12x + 9 = 0 are 3, -1 and -1. Explain why the fourth root must be a real

number. Find the fourth root.
Mathematics
1 answer:
LiRa [457]3 years ago
3 0
Descares Theorem says that the number of roots of a polynomial is equal to its highest exponent, and the roots could be real and/or imaginary (complex).

However is their is one complex root (a + i.b), it's conjugate is also
a root (a-i.b).

In the equation x⁴ - 4x³ - 2x² + 12x + 9 = 0, we have already 3 roots, then the 4rth cannot be a complex one since we don't have a fifth for its conjugate.

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Solving quadratic equations, it is found that he needs to charge:

1. He needs to charge $40 to break even.

2. He needs to charge $30 for a profit of $600.

<h3>What is a quadratic function?</h3>

A quadratic function is given according to the following rule:

y = ax^2 + bx + c

The solutions are:

  • x_1 = \frac{-b + \sqrt{\Delta}}{2a}
  • x_2 = \frac{-b - \sqrt{\Delta}}{2a}

In which:

\Delta = b^2 - 4ac

The profit equation in this problem is:

P(x) = -3x² + 150x - 1200.

He breaks even when P(x) = 0, hence:

-3x² + 150x - 1200 = 0.

The coefficients are a = -3, b = 150, c = -1200, hence:

  • \Delta = 150^2 - 4(-3)(-1200) = 8100
  • x_1 = \frac{-150 + \sqrt{8100}}{-6}
  • x_2 = \frac{-150 - \sqrt{8100}}{-6} = 40

He needs to charge $40 to break even.

For a profit of $600, we have that P(x) = 600, hence:

-3x² + 150x - 1200 = 600.

-3x² + 150x - 1800 = 0.

The coefficients are a = -3, b = 150, c = -1800, hence:

  • \Delta = 150^2 - 4(-3)(-1800) = 900
  • x_1 = \frac{-150 + \sqrt{900}}{-6}
  • x_2 = \frac{-150 - \sqrt{900}}{-6} = 30

He needs to charge $30 for a profit of $600.

More can be learned about quadratic equations at brainly.com/question/24737967

#SPJ1

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