I Need Help With a Pre-Calculus Questions

Approximate the real zeros of f(x) = 2x^4 - 3x^2 - 2 to the nearest tenth.

(a) -2

(b) + or - 1.4

(c) + or - 1.5

(d) no real zeros


Solve 2/x > -1/x-1

(a) x > 0

(b) x < 2/3, x > 1

(c) 0 < x < 1

(d) 0 < x < 2/3, x > 1

1. Factor as f(x) = (2x^2 + 1)(x^2 - 2)
This will only have real zeros for x^2 - 2 = 0, or x = ±√2 ≈ ±1.4, corresponding to (b).


2. Parentheses are needed around denominators.
.. 2/x > -1/(x-1)
.. 2(x-1) > -x ... multiply by x(x-1), assuming that product is > 0
.. 3x > 2
.. x > 2/3
This condition, together with x(x-1) > 0 means x > 1.

.. 2(x-1) < -x ... multiply by x(x-1), assuming that product is < 0
.. 3x < 2 ... add x+2
.. x < 2/3 ... divide by 3
This condition, together with x(x-1) < 0 means 0 < x < 2/3

The solution of the inequality is .. 0 < x < 2/3, x > 1, corresponding to (d).

For the first, substitute u=x^2 so the equation becomes:

2u^2 - 3u - 2 = 0

which is a quadratic in u, so use quadratic equation for roots:

u = [ 3 +/- sqrt( 9 + 16 ) ] / 4 = ( 3 +/- 5 ) / 4 = 2 or -1/2

The original x is therefore +/- sqrt(2), or approx +/-1.4. Note that u=-1/2 does not contribute any real solution for x.

For the second, if I'm reading your missing parentheses correctly:

2/x > -1/(x-1)

multiply both sides by x(x-1)

2(x-1) ?? -x

x ?? 2/3

Where the "??" means we must use ">" or "<" depending on whether the multiplier used above x(x-1) is positive or negative. But x(x-1) is positive for x<0 or x>1, hence

x > 2/3 if x<0 or x>1
x < 2/3 if 0 < x < 1

The first expression contributes x > 1 as the solution set, an the second contributes 0 < x < 2/3, hence the answer is (D)

IF YOU PROMISE NOT TO CUT MY HAIR.....
I will be glad to skype you through these