Solve the equations?

Question

1. In the polynomial function f(x)=x^3-4x^2+6x-4,
2 is a zero of the function. Name another zero of the function.
2.What value of x would be a restriction in the rational equation: 4/2x-3=4/5
3. Simplify:2y^2-2/3y^2+3y/2y-2/6y+2
4.solve equation showing work, list any restrictions on x: x/x+1=1/3+2/x-2
5.solve, listing any restrictions on x: 2/a=4/a^2

Answer 1

1. since we are finding the zeros of the function, we can set f(x) = 0 to get:
0 = x^3 - 4x² + 6x -4. Since zero times anything is zero, We separtate this poylnomial expression into 3 multiplying factors (each with a single x in it) and the value of the function is zero at each of the three values of x where one of the factors becomes zero. Thus, we can always write 0 = (x - a)( x - b)(x - c). The resulting zeros are x=a, x=b, and x=c.

Our problem now looks like
x^3 - 4x² + 6x - 4 = (x - 2)(x - b)(x- c) I have taken the liberty of substituting the original given function"zero": (x = 2) into our expression.

At this point is is easier to back up one step end factor the (x - 3) from th eoriginal. we do this by dividing. In condensed form it looks like (x^3 -4x² + 6x - 4) ÷( x - 2) = (x² - 2x + 2)

In long division it looks like:

.................x² - 2x + 2
............_______________
(x - 2) √ (x^3 -4x² + 6x - 4)
..............x^3 - 2x²
............______________
.....................-2x² + 6x
.....................-2x² + 4x
.............______________
...............................2x - 4
...............................2x - 4
.............................========

The last two zeros are harder to get. the polynomial x² - 2x + 2 cannot be factored using real numbers If you have not yet had imaginary numbers, then the answer to a second zero is where x² - 2x + 2 = 0 ( whereever that is). If you have had imaginary numbers, then use the quadratic equation to get the roots (which is another name for the zeros):
........................_________
.............-b ± √ (b² - 4ac)
.........x = .______________
..........................2a

which figures to x = 1 ± i for the other two zeros.


2. This one can be handled by inspection:

...............4............4
............_____.=.___
............2x - 3........5

This porblem is undefinied when the denominator of the fraction is zero. So, when x = 3/2 the problem is undefined and any conclusions drawn from it are untrustworthy! I once saw a proof the 2 = 3 in which every step was logical . The tricky problem was that at one point the denominator was zero (but it was "hidden" because the denominator used algebraic variables instead of just numbers in its expression).

It's getting late. I'll help you with the others tomorrow if you havn't already received valid answers from others

Answer 2

1. x^3-4x^2+6x-4 = 0
Divide by (x - 2) ......x^2 - 2x + 2
x^2 - 2x + 2 = 0
x^2 - 2x + 1 = -1 .... make the left hand a perfect square
x - 1 = i so x = (1 + i) is another root

2. : 4/(2x-3) = 4/5
Since you can not divide by 0, a restriction would be when 2x - 3 = 0 or x = 3/2

3. 2y^2 - 2/3y^2 + 3y/2y - 2/6
1/3y^2 + 3/2 - 2/6
1/3y^2 + 7/6

4. x/(x + 1) = 1/3 + 2/(x - 2)

X can not be -1 or 2 since either of these values will result in divide by zero.

Multiply through by the denominators (x + 1)*(x - 2)

x(x - 2) = (1/3)(x + 1)(x - 2) + 2(x + 1)
x^2 - 2x = (1/3)(x^2 - x - 2) + (2x + 2) = (1/3)x^2 + (5/3)x + 4/3
(2/3)x^2 - (11/3)x - 4/3 = 0
2x^2 - 11x - 4 = 0

x = (-b +/- SQRT(b^2 - 4ac))/2a
a = 2 and b = 11 and c = -4

x = (-11 +/- SQRT(121 - 16)/4
x = (-11 +/- SQRT(105))/4

5. 2/a=4/a^2
2a = 4
a = 2

There is no restriction on "a" since it is a constant

Answer 3

the answer is nineteen! distinctive how-to below: to make this extra convenient, group the x's including, of direction, their +/- indicators: 3x - 2x + 17 + 2 now that's an equation, and in case you notice no equivalent sign (=), then you definitely could anticipate that if there replaced into, the entire equation could be equivalent to 0: 3x - 2x + 17 + 2 = 0 now simplify-- use easy subtraction whilst coping with the variables (ignoring the x's, yet remembering to place them in the effect) this suggests which you're able to answer "3-2": 1x + 19 = 0 i've got been given the nineteen via including 17 and a couple of. you at the instant have a "1x", yet you may get rid of the "a million", with the aid of fact writing down merely "x" fairly skill "1x" x + 19 = 0 now to sparkling up it, the variable could be on one area, and the traditional fee on the different area. undergo in strategies that in case you're taking a huge decision or a variable around the equivalent sign, you're able to continuously exchange that's +/- sign! : x = -19 or 19 = -x now you have the answer! x is comparable to "detrimental 19" or "-19" or you could study it as "detrimental x is comparable to 19", it particularly is the comparable factor. desire this replaced into adequate element.