Find Interval(s) for Coefficient b

[nycmathguy]

Section 2.7
Question 88

Note:

Can you do 88 in step by step fashion as a guide for me to do a few more on my own?

Note:

We are done with Section 2.7. In fact, we are done with Chapter 2. Next stop: Chapter 3, Section 3.1 aka Exponential and Logarithmic Functions.

 

[MathLover1]

88'

a.

x^2+bx+9=0
x^2+bx+3^2=0
the equation has at least one real solution if its discriminant is >=0

b^2-4ac>=0 ...........in your case a=1, b=b, c=9

b^2-4*1*9>=0
b^2-36>=0
b^2>=36
b >=6
or
b >=-6

So, the possible intervals can be (-∞,-6) and (6,∞).

b.

Choose the first interval (-∞,-6) to test that whether the roots can be real or not

Substitute -6 for b into
b^2-36>=0
.(-6)^2-36≥0
.36-36≥0
0≥0 which is true, so the real root can be in the interval (-∞,-6)

Choose the first interval (6,∞) to test that whether the roots can be real or not

Substitute 6 for b into
b^2-36>=0
.(6)^2-36≥0
.36-36≥0
0≥0 which is true, so the real root can be in the interval (6,∞)

 

[nycmathguy]

88'

a.

x^2+bx+9=0
x^2+bx+3^2=0
the equation has at least one real solution if its discriminant is >=0

b^2-4ac>=0 ...........in your case a=1, b=b, c=9

b^2-4*1*9>=0
b^2-36>=0
b^2>=36
b >=6
or
b >=-6

So, the possible intervals can be (-∞,-6) and (6,∞).

b.

Choose the first interval (-∞,-6) to test that whether the roots can be real or not

Substitute -6 for b into
b^2-36>=0
.(-6)^2-36≥0
.36-36≥0
0≥0 which is true, so the real root can be in the interval (-∞,-6)

Choose the first interval (6,∞) to test that whether the roots can be real or not

Substitute 6 for b into
b^2-36>=0
.(6)^2-36≥0
.36-36≥0
0≥0 which is true, so the real root can be in the interval (6,∞)

Well-done. I will use your reply to do the remaining questions this weekend.

How about one more? See below.

Polynomial Function

 

[nycmathguy]

88'

a.

x^2+bx+9=0
x^2+bx+3^2=0
the equation has at least one real solution if its discriminant is >=0

b^2-4ac>=0 ...........in your case a=1, b=b, c=9

b^2-4*1*9>=0
b^2-36>=0
b^2>=36
b >=6
or
b >=-6

So, the possible intervals can be (-∞,-6) and (6,∞).

b.

Choose the first interval (-∞,-6) to test that whether the roots can be real or not

Substitute -6 for b into
b^2-36>=0
.(-6)^2-36≥0
.36-36≥0
0≥0 which is true, so the real root can be in the interval (-∞,-6)

Choose the first interval (6,∞) to test that whether the roots can be real or not

Substitute 6 for b into
b^2-36>=0
.(6)^2-36≥0
.36-36≥0
0≥0 which is true, so the real root can be in the interval (6,∞)

Question 90

Part (a)

2x^2 + bx + 5 = 0

Use b^2 - 4ac ≥ 0 to find b.

I found b to be ≥ -2•sqrt{10}.

Part (b)

Possible Intervals:

(-infinity, -2•sqrt{10} ) (2•sqrt{10}, infinity)

Use b^2 - 40 ≥ 0

Let b = -2sqrt{10}

(-2•sqrt{10})^2 - 40 ≥ 0.

This leads to 0 ≥ 0, which is true. The real root is found in the interval (-infinity, -2sqrt{10}).

Note: If my answer to 90 is right, it doesn't mean this question makes any sense to me at all. I have no idea what I did here.

 

[nycmathguy]

88'

a.

x^2+bx+9=0
x^2+bx+3^2=0
the equation has at least one real solution if its discriminant is >=0

b^2-4ac>=0 ...........in your case a=1, b=b, c=9

b^2-4*1*9>=0
b^2-36>=0
b^2>=36
b >=6
or
b >=-6

So, the possible intervals can be (-∞,-6) and (6,∞).

b.

Choose the first interval (-∞,-6) to test that whether the roots can be real or not

Substitute -6 for b into
b^2-36>=0
.(-6)^2-36≥0
.36-36≥0
0≥0 which is true, so the real root can be in the interval (-∞,-6)

Choose the first interval (6,∞) to test that whether the roots can be real or not

Substitute 6 for b into
b^2-36>=0
.(6)^2-36≥0
.36-36≥0
0≥0 which is true, so the real root can be in the interval (6,∞)

Question 90....A Continuation.

Let b = 2sqrt{10}

Plug into b^2 - 40 ≥ 0.

(2,sqrt{10})^2 - 40 ≥ 0

40 - 40 ≥ 0.

This leads to 0 ≥ 0, which is a true statement.
The real root can be found in the interval (2sqrt{10}, infinity).

I will now post 90 with my entire math work.

 

[nycmathguy]

Question 90 Completed

Part (a)

2x^2 + bx + 5 = 0

Use b^2 - 4ac ≥ 0 to find b.

I found b to be ≥ -2•sqrt{10}.

Part (b)

Possible Intervals:

(-infinity, -2•sqrt{10} ) (2•sqrt{10}, infinity)

Use b^2 - 40 ≥ 0

Let b = -2sqrt{10}

(-2•sqrt{10})^2 - 40 ≥ 0.

This leads to 0 ≥ 0, which is true. The real root is found in the interval (-infinity, -2sqrt{10}).

Let b = 2sqrt{10}

Plug into b^2 - 40 ≥ 0.

(2,sqrt{10})^2 - 40 ≥ 0

40 - 40 ≥ 0.

This leads to 0 ≥ 0, which is a true statement.
The real root can be found in the interval (2sqrt{10}, infinity).

Questions

1. What does all this mean?

2. What is the purpose of this exercise?

3. To me, this question is just teaching students how to find b in a polynomial. This is pointless. You say?

4. What conjecture is the question asking for?

 

[MathLover1]

Question 90 Completed

Part (a)

2x^2 + bx + 5 = 0

Use b^2 - 4ac ≥ 0 to find b.

I found b to be ≥ -2•sqrt{10}.

Part (b)

Possible Intervals:

(-infinity, -2•sqrt{10} ) (2•sqrt{10}, infinity)

Use b^2 - 40 ≥ 0

Let b = -2sqrt{10}

(-2•sqrt{10})^2 - 40 ≥ 0.

This leads to 0 ≥ 0, which is true. The real root is found in the interval (-infinity, -2sqrt{10}).

Let b = 2sqrt{10}

Plug into b^2 - 40 ≥ 0.

(2,sqrt{10})^2 - 40 ≥ 0

40 - 40 ≥ 0.

This leads to 0 ≥ 0, which is a true statement.
The real root can be found in the interval (2sqrt{10}, infinity).

Questions

1. What does all this mean?

2. What is the purpose of this exercise?

3. To me, this question is just teaching students how to find b in a polynomial. This is pointless. You say?

4. What conjecture is the question asking for?

1. What does all this mean?

how a conclusion can be formed on the basis of incomplete information

2. What is the purpose of this exercise?

Conjecture a relationship between the cycle decomposition of g and the characteristic polynomial of p(g)

3. To me, this question is just teaching students how to find b in a polynomial. This is pointless. You say?

in math nothing is pointless, you have to learn some shortcuts

4. What conjecture is the question asking for?

conjecture is an opinion or conclusion formed on the basis of incomplete information
asking you to conclude or suppose from grounds or evidence insufficient to ensure reliability

Conjectures
The graph of a polynomial of degree n has at most n x-intercepts.
The graph of a polynomial of odd degree has at least one x-intercept.
The graph of a polynomial of degree n has at most n-1 turning points.
The graph of a polynomial of even degree has at least one turning point.
The graph of a polynomial f(x)=a[n[*x^n+a[n-1]*x^(n-1)+....+a[0] behaves asymptotically like its leading term a[n]*x^n. That is, when x is large (approaching +∞ or -∞), f(x) has the same behaviour (approaching +∞ or -∞) as a[n]*x^n.

Another possible conjecture is that the graph of a polynomial of even degree has an odd number of turning points, while the graph of a polynomial of odd degree has an even number of turning points. Assuming the above conjectures, explain why this is true.

 

[nycmathguy]

1. What does all this mean?

how a conclusion can be formed on the basis of incomplete information

2. What is the purpose of this exercise?

Conjecture a relationship between the cycle decomposition of g and the characteristic polynomial of p(g)

3. To me, this question is just teaching students how to find b in a polynomial. This is pointless. You say?

in math nothing is pointless, you have to learn some shortcuts

4. What conjecture is the question asking for?

conjecture is an opinion or conclusion formed on the basis of incomplete information
asking you to conclude or suppose from grounds or evidence insufficient to ensure reliability

Conjectures
The graph of a polynomial of degree n has at most n x-intercepts.
The graph of a polynomial of odd degree has at least one x-intercept.
The graph of a polynomial of degree n has at most n-1 turning points.
The graph of a polynomial of even degree has at least one turning point.
The graph of a polynomial f(x)=a[n[*x^n+a[n-1]*x^(n-1)+....+a[0] behaves asymptotically like its leading term a[n]*x^n. That is, when x is large (approaching +∞ or -∞), f(x) has the same behaviour (approaching +∞ or -∞) as a[n]*x^n.

Another possible conjecture is that the graph of a polynomial of even degree has an odd number of turning points, while the graph of a polynomial of odd degree has an even number of turning points. Assuming the above conjectures, explain why this is true.

This is why you are a teacher and I am a joke.