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Sergio039 [100]
3 years ago
8

Solve the following C^2-14=5c

Mathematics
1 answer:
kipiarov [429]3 years ago
8 0

Answer:

c = - 2, c = 7

Step-by-step explanation:

Given

c² - 14 = 5c ( subtract 5c from both sides )

c² - 5c - 14 = 0 ← in standard form

To factorise the quadratic

Consider the factors of the constant term (- 14) which sum to give the coefficient of the c- term (- 5)

The factors are - 7 and + 2, since

- 7 × 2 = - 14 and - 7 + 2 = - 5, hence

(c - 7)(c + 2) = 0

Equate each factor to zero and solve for c

c - 7 = 0 ⇒ c = 7

c + 2 = 0 ⇒ c = - 2

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The function f(x) = x3 – 8x2 + x + 42 has zeros located at 7, –2, 3. Verify the zeros of f(x) and explain how you verified them.
xeze [42]

Answer:

  • zeros are {-2, 3, 7} as verified by graphing
  • end behavior: f(x) tends toward infinity with the same sign as x

Step-by-step explanation:

A graphing calculator makes finding or verifying the zeros of a polynomial function as simple as typing the function into the input box.

<h3>Zeros</h3>

The attachment shows the function zeros to be x ∈ {-2, 3, 7}, as required.

<h3>End behavior</h3>

The leading coefficient of this odd-degree polynomial is positive, so the value of f(x) tends toward infinity of the same sign as x when the magnitude of x tends toward infinity.

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<em>Additional comment</em>

The function is entered in the graphing calculator input box in "Horner form," which is also a convenient form for hand-evaluation of the function.

We know the x^2 coefficient is the opposite of the sum of the zeros:

  -(7 +(-2) +3) = -8 . . . . x^2 coefficient

And we know the constant is the opposite of the product of the zeros:

  -(7)(-2)(3) = 42 . . . . . constant

These checks lend further confidence that the zeros are those given.

(The constant is the opposite of the product of zeros only for odd-degree polynomials. For even-degree polynomials. the constant is the product of zeros.)

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