Prove all terms of quadratic pattern is positive.

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I recently did a Mathematics exam from a previous year, and I stumbled across a question's answer I struggled to fully understand.

It is given: The quadratic pattern 244 ; 193 ; 148 ; 109; ...

I've determined the $nth$ term as: $$Tn = 3n^2 -60n + 301$$ Now the questions asks, "Show that all the terms of the quadratic pattern is positive." Our teacher explained completing the square of the formula, but I couldn't catch what she said, and rather, I didn't understand why you would complete the square. I do however see that $3n^2 - 60n +301$ delivers ÃŽÂ” < 0 .

I can deduce that you would go in the direction of an inequality whereas you have $n >0$ (incomplete).

Perhaps someone with a higher understanding can explain this to me.

Thank you very much!

asked 1 hour ago

deadlyvirus

183

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up vote
2
down vote

favorite

I recently did a Mathematics exam from a previous year, and I stumbled across a question's answer I struggled to fully understand.

It is given: The quadratic pattern 244 ; 193 ; 148 ; 109; ...

I can deduce that you would go in the direction of an inequality whereas you have $n >0$ (incomplete).

Perhaps someone with a higher understanding can explain this to me.

Thank you very much!

asked 1 hour ago

deadlyvirus

183

New contributor

add a commentÂ |Â

up vote
2
down vote

favorite

I recently did a Mathematics exam from a previous year, and I stumbled across a question's answer I struggled to fully understand.

It is given: The quadratic pattern 244 ; 193 ; 148 ; 109; ...

I can deduce that you would go in the direction of an inequality whereas you have $n >0$ (incomplete).

Perhaps someone with a higher understanding can explain this to me.

Thank you very much!

asked 1 hour ago

deadlyvirus

183

New contributor

I recently did a Mathematics exam from a previous year, and I stumbled across a question's answer I struggled to fully understand.

It is given: The quadratic pattern 244 ; 193 ; 148 ; 109; ...

I can deduce that you would go in the direction of an inequality whereas you have $n >0$ (incomplete).

Perhaps someone with a higher understanding can explain this to me.

Thank you very much!

quadratics

asked 1 hour ago

deadlyvirus

183

New contributor

asked 1 hour ago

deadlyvirus

183

New contributor

asked 1 hour ago

deadlyvirus

183

New contributor

asked 1 hour ago

deadlyvirus

183

asked 1 hour ago

deadlyvirus

183

New contributor

deadlyvirus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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3 Answers
3

active

oldest

votes

up vote
2
down vote

accepted

Completing the square gives you $ Delta leq 0$

It shows that equation will always be positive.

But, it is easier to do so using calculus.

Take $$y=3n^2-60n+301$$

Take first derivative with respect to n: $$fracdydn=y'=6n-60$$
$$y'=6(n-10)$$
As you can see, for all $$n <10 , y'< 0$$

And, for all $$n>10, y'>0$$

It means that function $y=f(x)$ attains its minimum at x=10. f(10)=1

It decreases before $x=10$ and increases after $x=10$.

So, it can never be negative.

have a look at it's graph

edited 48 mins ago

answered 53 mins ago

omega

1,1452818

Thank you very much @omega, this made it very clear, may I say I do feel quite stupid after viewing it. Thank you! The calculus really made the difference!
â€“Â deadlyvirus
22 mins ago

add a commentÂ |Â

up vote
2
down vote

First, we know all squares are positive.

$$(n-10)^2 ge 0\
3(n-10)^2 ge 0 \
3(n-10)^2 +1 ge 1 >0 \
3(n-10)^2 +1 >0 $$

By expanding we will get
$$3n^2-60n+301>0$$

The way you should approach this is by thinking: what must be positive? squares. Then, how do I come up with a square? Completing the Square.

Completing the Square is explained well here.

answered 57 mins ago

Vee Hua Zhi

79119

Thank you @Vee Hau Zhi for your comment, it really did help me to understand. And as I said in omega's post, I feel quite stupid after seeing both your answers. It's not completing the square I had problems with, it was why it was necessary for this question. But as I said, I do feel quite embarrassed, but nevertheless, your answer helped. Thank you very much!
â€“Â deadlyvirus
18 mins ago

add a commentÂ |Â

up vote
1
down vote

Completing the square enable you to see the vertex clearly. For $f(x)=ax^2+bx+c$ where $a>0$, we can see the minimal value that it can attain.

Alternatively, just see that the discriminant is negative, that is the function doesn't intersect the $x$-axis and it doesn't change sign. Since one of the term is positive, every term is positive.

answered 55 mins ago

Siong Thye Goh

87.9k1460111

Ah, the graphing terms along with the graph link from omega, I understood the overall picture more clearly. Thank you very much!
â€“Â deadlyvirus
16 mins ago

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
2
down vote

accepted

Completing the square gives you $ Delta leq 0$

It shows that equation will always be positive.

But, it is easier to do so using calculus.

Take $$y=3n^2-60n+301$$

Take first derivative with respect to n: $$fracdydn=y'=6n-60$$
$$y'=6(n-10)$$
As you can see, for all $$n <10 , y'< 0$$

And, for all $$n>10, y'>0$$

It means that function $y=f(x)$ attains its minimum at x=10. f(10)=1

It decreases before $x=10$ and increases after $x=10$.

So, it can never be negative.

have a look at it's graph

edited 48 mins ago

answered 53 mins ago

omega

1,1452818

Thank you very much @omega, this made it very clear, may I say I do feel quite stupid after viewing it. Thank you! The calculus really made the difference!
â€“Â deadlyvirus
22 mins ago

add a commentÂ |Â

up vote
2
down vote

accepted

Completing the square gives you $ Delta leq 0$

It shows that equation will always be positive.

But, it is easier to do so using calculus.

Take $$y=3n^2-60n+301$$

Take first derivative with respect to n: $$fracdydn=y'=6n-60$$
$$y'=6(n-10)$$
As you can see, for all $$n <10 , y'< 0$$

And, for all $$n>10, y'>0$$

It means that function $y=f(x)$ attains its minimum at x=10. f(10)=1

It decreases before $x=10$ and increases after $x=10$.

So, it can never be negative.

have a look at it's graph

edited 48 mins ago

answered 53 mins ago

omega

1,1452818

Thank you very much @omega, this made it very clear, may I say I do feel quite stupid after viewing it. Thank you! The calculus really made the difference!
â€“Â deadlyvirus
22 mins ago

add a commentÂ |Â

up vote
2
down vote

accepted

Completing the square gives you $ Delta leq 0$

It shows that equation will always be positive.

But, it is easier to do so using calculus.

Take $$y=3n^2-60n+301$$

Take first derivative with respect to n: $$fracdydn=y'=6n-60$$
$$y'=6(n-10)$$
As you can see, for all $$n <10 , y'< 0$$

And, for all $$n>10, y'>0$$

It means that function $y=f(x)$ attains its minimum at x=10. f(10)=1

It decreases before $x=10$ and increases after $x=10$.

So, it can never be negative.

have a look at it's graph

edited 48 mins ago

answered 53 mins ago

omega

1,1452818

Completing the square gives you $ Delta leq 0$

It shows that equation will always be positive.

But, it is easier to do so using calculus.

Take $$y=3n^2-60n+301$$

Take first derivative with respect to n: $$fracdydn=y'=6n-60$$
$$y'=6(n-10)$$
As you can see, for all $$n <10 , y'< 0$$

And, for all $$n>10, y'>0$$

It means that function $y=f(x)$ attains its minimum at x=10. f(10)=1

It decreases before $x=10$ and increases after $x=10$.

So, it can never be negative.

have a look at it's graph

edited 48 mins ago

answered 53 mins ago

omega

1,1452818

edited 48 mins ago

answered 53 mins ago

omega

1,1452818

answered 53 mins ago

omega

1,1452818

answered 53 mins ago

omega

1,1452818

Thank you very much @omega, this made it very clear, may I say I do feel quite stupid after viewing it. Thank you! The calculus really made the difference!
â€“Â deadlyvirus
22 mins ago

add a commentÂ |Â

Thank you very much @omega, this made it very clear, may I say I do feel quite stupid after viewing it. Thank you! The calculus really made the difference!
â€“Â deadlyvirus
22 mins ago

Thank you very much @omega, this made it very clear, may I say I do feel quite stupid after viewing it. Thank you! The calculus really made the difference!
â€“Â deadlyvirus
22 mins ago

add a commentÂ |Â

up vote
2
down vote

First, we know all squares are positive.

$$(n-10)^2 ge 0\
3(n-10)^2 ge 0 \
3(n-10)^2 +1 ge 1 >0 \
3(n-10)^2 +1 >0 $$

By expanding we will get
$$3n^2-60n+301>0$$

The way you should approach this is by thinking: what must be positive? squares. Then, how do I come up with a square? Completing the Square.

Completing the Square is explained well here.

answered 57 mins ago

Vee Hua Zhi

79119

Thank you @Vee Hau Zhi for your comment, it really did help me to understand. And as I said in omega's post, I feel quite stupid after seeing both your answers. It's not completing the square I had problems with, it was why it was necessary for this question. But as I said, I do feel quite embarrassed, but nevertheless, your answer helped. Thank you very much!
â€“Â deadlyvirus
18 mins ago

add a commentÂ |Â

up vote
2
down vote

First, we know all squares are positive.

$$(n-10)^2 ge 0\
3(n-10)^2 ge 0 \
3(n-10)^2 +1 ge 1 >0 \
3(n-10)^2 +1 >0 $$

By expanding we will get
$$3n^2-60n+301>0$$

The way you should approach this is by thinking: what must be positive? squares. Then, how do I come up with a square? Completing the Square.

Completing the Square is explained well here.

answered 57 mins ago

Vee Hua Zhi

79119

Thank you @Vee Hau Zhi for your comment, it really did help me to understand. And as I said in omega's post, I feel quite stupid after seeing both your answers. It's not completing the square I had problems with, it was why it was necessary for this question. But as I said, I do feel quite embarrassed, but nevertheless, your answer helped. Thank you very much!
â€“Â deadlyvirus
18 mins ago

add a commentÂ |Â

up vote
2
down vote

First, we know all squares are positive.

$$(n-10)^2 ge 0\
3(n-10)^2 ge 0 \
3(n-10)^2 +1 ge 1 >0 \
3(n-10)^2 +1 >0 $$

By expanding we will get
$$3n^2-60n+301>0$$

The way you should approach this is by thinking: what must be positive? squares. Then, how do I come up with a square? Completing the Square.

Completing the Square is explained well here.

answered 57 mins ago

Vee Hua Zhi

79119

First, we know all squares are positive.

$$(n-10)^2 ge 0\
3(n-10)^2 ge 0 \
3(n-10)^2 +1 ge 1 >0 \
3(n-10)^2 +1 >0 $$

By expanding we will get
$$3n^2-60n+301>0$$

The way you should approach this is by thinking: what must be positive? squares. Then, how do I come up with a square? Completing the Square.

Completing the Square is explained well here.

answered 57 mins ago

Vee Hua Zhi

79119

answered 57 mins ago

Vee Hua Zhi

79119

answered 57 mins ago

Vee Hua Zhi

79119

answered 57 mins ago

Vee Hua Zhi

79119

Thank you @Vee Hau Zhi for your comment, it really did help me to understand. And as I said in omega's post, I feel quite stupid after seeing both your answers. It's not completing the square I had problems with, it was why it was necessary for this question. But as I said, I do feel quite embarrassed, but nevertheless, your answer helped. Thank you very much!
â€“Â deadlyvirus
18 mins ago

add a commentÂ |Â

Thank you @Vee Hau Zhi for your comment, it really did help me to understand. And as I said in omega's post, I feel quite stupid after seeing both your answers. It's not completing the square I had problems with, it was why it was necessary for this question. But as I said, I do feel quite embarrassed, but nevertheless, your answer helped. Thank you very much!
â€“Â deadlyvirus
18 mins ago

Thank you @Vee Hau Zhi for your comment, it really did help me to understand. And as I said in omega's post, I feel quite stupid after seeing both your answers. It's not completing the square I had problems with, it was why it was necessary for this question. But as I said, I do feel quite embarrassed, but nevertheless, your answer helped. Thank you very much!
â€“Â deadlyvirus
18 mins ago

add a commentÂ |Â

up vote
1
down vote

Completing the square enable you to see the vertex clearly. For $f(x)=ax^2+bx+c$ where $a>0$, we can see the minimal value that it can attain.

Alternatively, just see that the discriminant is negative, that is the function doesn't intersect the $x$-axis and it doesn't change sign. Since one of the term is positive, every term is positive.

answered 55 mins ago

Siong Thye Goh

87.9k1460111

Ah, the graphing terms along with the graph link from omega, I understood the overall picture more clearly. Thank you very much!
â€“Â deadlyvirus
16 mins ago

add a commentÂ |Â

up vote
1
down vote

Completing the square enable you to see the vertex clearly. For $f(x)=ax^2+bx+c$ where $a>0$, we can see the minimal value that it can attain.

Alternatively, just see that the discriminant is negative, that is the function doesn't intersect the $x$-axis and it doesn't change sign. Since one of the term is positive, every term is positive.

answered 55 mins ago

Siong Thye Goh

87.9k1460111

Ah, the graphing terms along with the graph link from omega, I understood the overall picture more clearly. Thank you very much!
â€“Â deadlyvirus
16 mins ago

add a commentÂ |Â

up vote
1
down vote

Completing the square enable you to see the vertex clearly. For $f(x)=ax^2+bx+c$ where $a>0$, we can see the minimal value that it can attain.

Alternatively, just see that the discriminant is negative, that is the function doesn't intersect the $x$-axis and it doesn't change sign. Since one of the term is positive, every term is positive.

answered 55 mins ago

Siong Thye Goh

87.9k1460111

Completing the square enable you to see the vertex clearly. For $f(x)=ax^2+bx+c$ where $a>0$, we can see the minimal value that it can attain.

Alternatively, just see that the discriminant is negative, that is the function doesn't intersect the $x$-axis and it doesn't change sign. Since one of the term is positive, every term is positive.

answered 55 mins ago

Siong Thye Goh

87.9k1460111

answered 55 mins ago

Siong Thye Goh

87.9k1460111

answered 55 mins ago

Siong Thye Goh

87.9k1460111

answered 55 mins ago

Siong Thye Goh

87.9k1460111

Ah, the graphing terms along with the graph link from omega, I understood the overall picture more clearly. Thank you very much!
â€“Â deadlyvirus
16 mins ago

add a commentÂ |Â

Ah, the graphing terms along with the graph link from omega, I understood the overall picture more clearly. Thank you very much!
â€“Â deadlyvirus
16 mins ago

Ah, the graphing terms along with the graph link from omega, I understood the overall picture more clearly. Thank you very much!
â€“Â deadlyvirus
16 mins ago

add a commentÂ |Â

deadlyvirus is a new contributor. Be nice, and check out our Code of Conduct.

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