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Dovator [93]
3 years ago
8

How do you solve x^4 - 3x^3 - 3x^2 - 75x - 700

Mathematics
1 answer:
madam [21]3 years ago
4 0

Answer:

  = (x +4)(x -7)(x^2 +25)

  roots are -4, 7, ±5i

Step-by-step explanation:

You have not said what "solve" means in this context. An expression by itself doesn't have a solution. We have assumed you want to find the factoring and/or roots of it.

I like to use a graphing calculator to find the real roots. For this expression, there are two of them, one positive and one negative. (You know there will be one positive real root, and at least one negative real root, from Descartes's rule of signs.)

Then those roots can be factored out and the solution to the remaining quadratic determined. That factoring can occur by polynomial long division, synthetic division, or other means.

I like to see what happens when I plot the graph of the function divided by the known factors. (We expect a parabola that doesn't cross the x-axis.) The vertex of that parabola can be used to find the remaining roots.

The x-intercepts of the given expression are -4 and +7, so two of the factors are (x+4) and (x-7). Dividing these from the given expression (by synthetic division or other means) gives (x^2 +25). This only has imaginary roots (±5i).

____

If you're constrained to doing this "by hand" with only a scientific calculator, Descartes's rule of signs tells you there is one positive real root. (Only one sign change in the sequence of coefficient signs: +----.)

The rational root theorem tells you it will be a divisor of 700. Various estimates of the maximum magnitude of it will tell you it is probably less than 14 (easily checked). So, the numbers you can test as roots would be 1, 2, 4, 5, 7, 10, 14. You will find that 7 is a root, and then you can reduce the problem to the cubic x^3 +4x^2 +25x +100.

When odd-degree term signs are changed, there will be 3 sign changes (-+-+), hence at least one negative real root. The rational root theorem tells you it is a divisor of 100, so possible choices are -1, -2, -4, -5. By trial and error or other means, you can find the root to be -4. Then the problem reduces to the quadratic x^2 +25.

Roots of that are ±√(-25) = ±5i.

This process generally entails a fair amount of trial-and-error work, which is why I prefer one that makes some use of technology.

_____

We have presumed you have some familiarity with ...

  • Descartes's rule of signs
  • Rational Root Theorem
  • synthetic division

This will usually be the case when you're presented with problems like this. If you need additional information on any of these, it is readily available on the internet (and probably also in your reference material).

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Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:
Papessa [141]

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

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This clearly doesn't satisfy the differential equation, hence, it is not a solution.

6 0
3 years ago
CORRECT ANSWER GETS BRAINLIESTTTT!!!!!
VladimirAG [237]

Senior tickets (x)

Child tickets (y)

First day:  3x + 5y = 70

second day: 12x + 12y = 216

Solve the system of equations (use elimination)

Multiply first equation by -4   -4(3x + 5y = 70),  which makes it

-12x - 20y = -280

(+)12x + 12y = 216    add to second equation

-8y = -64

divide by -8.    y = 8

Plugin the y value to either equation ( I will choose first equation)

3x + 5(8) =70

3x+ 40 = 70

3x = 30

x = 10

Senior tickets are $10, child tickets are $8

7 0
3 years ago
PLEASE CHECK MY ANSWER (Will mark brainliest)
mixer [17]
I believe you are correct
5 0
3 years ago
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Elan Coil [88]
<span>solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable.</span>
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Which describes how to calculate the range of this data set?
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Step-by-step explanation:

For the range you subtract the smallest number from the biggest number so 12-4.

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