Solve algebraically for all values of x: -4x^3+2x^2+8x-4=0

6 answers
Question:

Solve algebraically for all values of x: -4x^3+2x^2+8x-4=0

Answers

1) cross multiply both sides.
3x^2-24= 2x^2+10
x^2-10x-24=0
x^2-12x+2x-24=0
x(x-12)+2(x-12)=0
(x-12) (x+2)=0
Bring -12 and 2 to the other side. Therefore the value of x are 12 and -2.

9.17

9.78

Step-by-step explanation:

Solution:-

a)

The following information is provided:

Case 1:

- Service rate ( μ ) = 12 customer per hour ( 60 minutes per hour / 5 minutes time per customer )

- Arrival rate ( λ ) = 10 per hour

- Number of servers ( s ) = 2

- Initially determine the value of utilization ( ρ ):

             ρ = λ / s*μ

             ρ = 10 / 2*12

             ρ = 0.4167 - 41.67%  

- Determine the values of L, Lq , W and Wq. using the given formulas in the spreadsheet attached. Figure 1

- Figure 2 expresses the result of all the calculations.

Case 2:

- Service rate ( μ ) = 12 customer per hour ( 60 minutes per hour / 5 minutes time per customer )

- Arrival rate ( λ ) = 5 per hour

- Number of servers ( s ) = 1

- Initially determine the value of utilization ( ρ ):

             ρ = λ / s*μ

             ρ = 5 / 1*12

             ρ = 0.4167 - 41.67%  

- Determine the values of L, Lq , W and Wq. using the given formulas in the spreadsheet attached. Figure 3, expresses the results

- Comparing the outputs of both the cases we see that next year's system yields smaller values of L; however, the values of Lq, W ,Wq are higher for next year as compared to last years.

W for the current year = 0.24

Mean arrival rate ( λ ) = 5 per hour

Number of servers ( s ) = 1

- Determine the value of service rates ( μ ) that would provide W = 0.24 hrs. Use the given formula:

           W = 1 / ( μ - λ )

           μ = 1 / W + λ

           μ = 1 /0.24 + 5

           μ = 9.17 customers per hour

- Hence, its concluded that the required service rate is 9.17 customers per hour.

c)

Consider the value of μ is adjustable now. Now find the value of μ such that it takes the value of "Wq" from current year.

 Wq for the current year = 0.10667 hours

 Mean arrival rate ( λ ) = 5 per hour

 Number of servers ( s ) = 1

- Determine the value of service rates ( μ ) that would provide Wq = 0.10667 hrs. Use the given formula:

             Wq = λ / μ*( μ - λ )

             Wq*μ^2 - Wq*λ*μ - λ = 0

             0.10667*μ^2 - 0.53335*μ - 5 = 0

- Solve the quadratic and obtain values of (μ):

             μ = 9.78 , -4.78

- The negative value of μ is discarded while the positive value μ = 9.78 customers per hours is the required service rate.


[tex]People's Software Company has just set up a call center to provide technical assistance on its new s[/tex]
[tex]People's Software Company has just set up a call center to provide technical assistance on its new s[/tex]
[tex]People's Software Company has just set up a call center to provide technical assistance on its new s[/tex]

The values of 'x' are -1.2, 0, 0, [tex]-4i[/tex] or [tex]4i[/tex].

Step-by-step explanation:

Given:

The equation to solve is given as:

[tex]5x^5+6x^4+80x^3+96x^2=0[/tex]

Factoring [tex]x^2[/tex] from all the terms, we get:

[tex]x^2(5x^3+6x^2+80x+96)=0[/tex]

Now, rearranging the terms, we get:

[tex]x^2(5x^3+80x+6x^2+96)=0[/tex]

Now, factoring [tex]5x[/tex] from the first two terms and 6 from the last two terms, we get:

[tex]x^2(5x(x^2+16)+6(x^2+16))=0\\x^2(x^2+16)(5x+6)=0[/tex]

Now, equating each factor to 0 and solving for 'x', we get:

[tex]x^2=0\\x=0\ and\ 0\\\\5x+6=0\\x=\frac{-6}{5}=1.2\\\\x^2+16=0\\x^2=-16\\x=\sqrt{-16}=\pm 4i[/tex]

There are 3 real values and 2 imaginary values. The value of 'x' as 0 is repeated twice.

Therefore, the values of 'x' are -1.2, 0, 0, [tex]-4i[/tex] or [tex]4i[/tex].

x=5

sqrt of 5-4=1

then 1 +x(5)=6

X = -1.949844

81 2(-1.949844)³ + (-1.949844)² = 0
27 6(-1.949844) - 3 = 0 
0 = 0

X-4+x=6
Fill in x=5
5-4=1+5= 6
Therefore, the value of x is 5.

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