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finlep [7]
3 years ago
6

What is the possible number of positive and negative real roots or zeroes in f(x)=x^3+x^2-4

Mathematics
1 answer:
Tanya [424]3 years ago
3 0

This is a third degree polynomial, so we know that the maximum amount of roots will be 3.

We also know that imaginary roots come in pairs, so there is either 0 or 2 imaginary roots.

We can tell the amount of positive / negative roots by applying Descartes' rule of signs.

Let's examine the polynomial: x^3 + x^2 - 4

Note the sign changes.

From left to right, there is one sign change.

We now know there is only one positive real root.

Now, we can replace all values of x with -1 and then simplify. The number of sign changes will tell us the amount of negative roots.

(-1)^3 + (-1)^2 - 4

-1 + 1 - 4

From left to right, there are two sign changes, which tells us there will be 2 or 0 negative roots.

If there are 2, they are not imaginary.

If there are 0, they are imaginary.

<h3>So, our guaranteed root is:</h3><h3>1 positive, real root.</h3><h3>And our possible roots are:</h3><h3>2 negative, real roots.</h3><h3>0 negative, real roots, but 2 imaginary roots.</h3>
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We have been given an image of a circle. We are asked to find the value of each variable.

We can see that angle b corresponds to diameter of circle. We know that the measure of an angle that is inscribed to the diameter of a circle is 90 degrees . Therefore, the value of b will be 90.

we can see that angle is an inscribed angle of arc 99 degrees. We know that measure of an inscribed angle is half the measure of inscribed arc.

a=\frac{1}{2}\cdot 99

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Therefore, the value of a is 49.5 units.

We know that measure of all angles of a triangle is 180 degrees. The measure of 3rd angle will be half the measure of c, so we can set an equation as:

a+b+\frac{1}{2}c=180

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Therefore, the value of c is 81.

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I really don’t get this , it’s graphing linear inequalities
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ILL GIVE BRAINLIEST (EASY STUFF)
insens350 [35]
7 multiplied by 11 is 77. And to get 462, (and based on the given information), we can conclude that it is 3*2*7*11.
3 0
3 years ago
Read 2 more answers
3,6,11,18,27,38
MakcuM [25]

Answer:  The pattern is x^2+2

where x is the term number

Example: the 5th term is 27 because x = 5 leads to x^2+2 = 5^2+2 = 27

=================================================

Explanation:

  • The jump from 3 to 6 is +3
  • The jump from 6 to 11 is +5
  • The jump from 11 to 18 is +7
  • The jump from 18 to 27 is +9
  • The jump from 27 to 38 is +11

The pattern of jumps is: 3, 5, 7, 9, 11

Those increments are going up by 2 each time.

Since we have a consistent pattern of increments, this means that the sequence follows a quadratic model.

Quadratics are stuff like x^2+7x+10 or 3x^2-7. The leading term has an exponent of 2.

-----------

If x is the term number and y is the term itself, then we have these points

(1,3)

(2,6)

(3,11)

(4,18)

(5,27)

(6,38)

The x coordinates increase by 1 each time. The y coordinates are the terms given by your teacher.

Pick exactly 3 of those points. I'll pick the first 3.

Why 3? Because we'll have 3 unknowns to solve for, in which we'll need 3 equations.

  • Plug (x,y) = (1,3) into y = ax^2+bx+c, then simplify. You should get the equation a+b+c = 3
  • Repeat for (x,y) = (2,6) and you should get 4a+2b+c = 6
  • Repeat for (x,y) = (3,11) and you should get 9a+3b+c = 11

-----------

We have this system of equations

\begin{cases}a+b+c = 3\\ 4a+2b+c = 6\\9a+3b+c = 11\end{cases}

There are a number of methods to solve this system. Substitution is what I'll go for.

Solve the first equation for c

a+b+c = 3

c = 3-a-b

Then use substitution.

4a+2b+c = 6

4a+2b+(3-a-b) = 6

3a+b+3 = 6

3a+b = 6-3

3a+b = 3

and

9a+3b+c = 11

9a+3b+(3-a-b) = 11

8a+2b + 3 = 11

8a+2b = 11-3

8a+2b = 8

We now have this reduced system of equations.

\begin{cases}3a+b = 3\\8a+2b = 8\end{cases}

I'll skip the steps as this solution is getting very lengthy as it is. The basic idea is to use substitution again. You should find that a = 1 and b = 0 form the solution set here.

Use those values to find c

c = 3-a-b

c = 3-1-0

c = 2

-----------

To summarize the previous section, we have these solutions:

a = 1, b = 0, c = 2

Therefore the equation y = ax^2+bx+c becomes y = 1x^2+0x+2 aka y = x^2+2. This lets us find any term.

Let's test it out.

  • If x = 1, then y = x^2+2 = 1^2+2 = 3
  • If x = 2, then y = x^2+2 = 2^2+2 = 6
  • If x = 3, then y = x^2+2 = 3^2+2 = 11

And so on. I'll let you test the other x values (4 through 6).

Another way to confirm the answer is to subtract 2 from each item in the original set {3,6,11,18,27,38} and you'll end up with {1,4,9,16,25,36}. This is the list of perfect squares. It shows that term x is simply x^2 but add on 2 so things are adjusted accordingly.

Side note: you can use a tool like GeoGebra or WolframAlpha to quickly solve the system of equations. However, I recommend it only as a means to check your answer rather than do the work for you.

7 0
2 years ago
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