Mathematics, 04.07.2019 05:30 harleyloveless2853

Could someone explain these to me and answer them? you!

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Answer from: westlakebuddy7884

These are two questions and two answers.

Question 1. For all x in the domain, the function is equivalent to:

$J.\frac{1}{x^2-x}$

Explanation:

First note the the domain of the function is all the real values except those for which the denominator is zero, this is x³ - x = 0.

And x³ - x = 0 ⇒ x (x² - 1) = 0 ⇒ x (x - 1) (x + 1) = 0

⇒ x = 0, x = 1, x = - 1.

∴ The domain is all the real values such that x ≠0, x ≠ - 1, and x ≠ 1.

Now, you can simplify the function following these steps:

$\frac{x+1}{x^3-x} = \frac{x+1}{x(x^2-1)}=\frac{x+1}{x(x+1)(x-1)}= \frac{1}{x(x-1)}= \frac{1}{x^2-x}$

So, the answer is the option J.

Question 2. What are the real solutions to the equation |x|² + 2|x| - 3 = 0?

option F. +/- 1

Explanation:

1) Use the fact that |x|² = x²

⇒ x² + 2|x| - 3 = 0

2) Transpose terms to isolate 2|x|:

⇒2|x| = 3 - x²

3) Square both sides:

⇒ [2|x|] ² = (3 - x² )²

⇒ 4|x|² = 9 - 6x² + x⁴

⇒ 4x² = 9 -6x² + x⁴

4) Transpose terms:

⇒ x⁴ - 4x² - 6x² + 9 = 0

⇒ x⁴ - 10x² + 9 = 0

5) Change variable: x² = u

⇒ u² - 10u + 9 = 0

6) Factor:

⇒ (u - 9) (u - 1) = 0

u - 9 = 0 ⇒ u = 9u - 1 = 0 ⇒ u = 1

6) Comeback to the variable x (undo the change of variable):

⇒ x² = u ⇒

x² = 9 ⇒ x = 3, x = - 3x² = 1 ⇒ x = 1, x = - 1

6) Verify extraneous solutions:

|x|² + 2|x| - 3 = 0

Replace x = 3

⇒ |3|² + 2|3| - 3 = 0

⇒ 9 + 6 - 3 = 0

⇒ 12 = 0 ⇒ FALSE ⇒ extraneous solution ⇒ discarded

Replace x = - 3

⇒ |-3|² + 2|-3| - 3 = 0

⇒ 9 + 6 - 3 = 0

⇒ 12 = 0 ⇒ FALSE ⇒ extraneous solution ⇒ discarded

Replace x = 1

⇒ |1|² + 2|1| - 3 = 0

⇒ 1 + 2 - 3 = 0

⇒ 0 = 0 ⇒ TRUE ⇒ actual solution

Replace x = - 1

⇒ |-1|² + 2|-1| - 3 = 0

⇒ 1 + 2 - 3 = 0

⇒ 0 = 0 ⇒ TRUE ⇒ actual solution

Conclusion: the solutions are +1 and -1.

Answer from: Quest

The first answer is correct
Asap plz i have 7 min left on the test

Answer from: Quest

The equation is y=17(7)+600 and it equals 717

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