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3 years ago

Need help with this math problem

Mathematics

2 answers:

MissTica3 years ago

8 0

Answer:

Please ignore - I missed the part about coefficients must be real and did not account for the conjugate root 7+i. Sorry!

Step-by-step explanation:

Three roots, 1-sqrt3, 1+sqrt3 and 7-i, are given so the polynomial is of third degree.

To have leading coefficient of 1, simply the term (x-root) together for given roots:

P(x) = (x-(1-sqrt3)) * ((x-(1+sqrt3)) * (x-(7-i))

= (x^2-(1-sqrt3+1+sqrt3)x+(1-sqrt3)(1+sqrt3)) * (x-7+i)

= (x^2-2x+1-3) * (x-7+i)

= (x^2-2x-2) * (x-7+i)

= x^3 - (9-i)x^2 + (12-2i)x + (14-2i)

if x is real, then it can be re-arranged to real and imaginary term:

P(x) = x^3 - 9x^2 + 12x + 14 + (x^2 - 2 x - 2)i

makvit [3.9K]3 years ago

7 0

Answer:

P(x) = x^4 -16x^3 +76x^2 -72x -100

Step-by-step explanation:

The two roots 1-√3 and 1+√3 give rise to the quadratic factor ...

... (x -(1-√3))(x -(1+√3)) = (x-1)^2 -(√3)^2 = x^2 -2x -2

The complex root 7-i has a conjugate that is also a root. These two roots give rise to the quadratic factor ...

... (x -(7 -i))(x -(7 +i)) = (x-7)^2 -(i)^2 = x^2 -14x +50

The product of these two quadratic factors is ...

... P(x) = (x^2 -2x -2)(x^2 -14x +50) = x^4 +x^3(-14 -2) +x^2(50 +28 -2) +x(-100+28) -100

... P(x) = x^4 -16x^3 +76x^2 -72x -100

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Which of the following is the result of the equation below after completing the square and factoring? x^2+3x+8=6

Inessa05 [86]

Answer:

Step-by-step explanation:

6 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals: