Let $a in mathbb R$. Prove that ($x^2 + ax + a > 0$ for every $x in mathbb R)$ iff $(0< a<4)$...

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This is a homework question for my university math proofs course. I am asked to prove the above bi conditional statement. The text book gives the hint that completing the square of the quadratic equation will simplify it.

Thanks,
John

proof-writing quadratics

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edited Jan 25 at 5:00

Andrei

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asked Jan 25 at 4:50

John TophamJohn Topham

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closed as off-topic by Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos Jan 25 at 13:09

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• 
  
  
  

  
  

  
  
  
  
  
  

  
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  So, what happens when you do complete the square?
  $endgroup$
  – Lord Shark the Unknown
  Jan 25 at 4:51
  

  
  
  
  

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-1

$begingroup$

1. List item

This is a homework question for my university math proofs course. I am asked to prove the above bi conditional statement. The text book gives the hint that completing the square of the quadratic equation will simplify it.

Thanks,
John

proof-writing quadratics

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edited Jan 25 at 5:00

Andrei

13.1k21230

asked Jan 25 at 4:50

John TophamJohn Topham

63

$endgroup$

closed as off-topic by Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos Jan 25 at 13:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, what happens when you do complete the square?
  $endgroup$
  – Lord Shark the Unknown
  Jan 25 at 4:51
  

  
  
  
  

add a comment | 

-1

-1

-1

$begingroup$

1. List item

This is a homework question for my university math proofs course. I am asked to prove the above bi conditional statement. The text book gives the hint that completing the square of the quadratic equation will simplify it.

Thanks,
John

proof-writing quadratics

share|cite|improve this question

edited Jan 25 at 5:00

Andrei

13.1k21230

asked Jan 25 at 4:50

John TophamJohn Topham

63

$endgroup$

1. List item

This is a homework question for my university math proofs course. I am asked to prove the above bi conditional statement. The text book gives the hint that completing the square of the quadratic equation will simplify it.

Thanks,
John

proof-writing quadratics

proof-writing quadratics

share|cite|improve this question

edited Jan 25 at 5:00

Andrei

13.1k21230

asked Jan 25 at 4:50

John TophamJohn Topham

63

share|cite|improve this question

edited Jan 25 at 5:00

Andrei

13.1k21230

asked Jan 25 at 4:50

John TophamJohn Topham

63

share|cite|improve this question

share|cite|improve this question

edited Jan 25 at 5:00

Andrei

13.1k21230

edited Jan 25 at 5:00

Andrei

13.1k21230

edited Jan 25 at 5:00

Andrei

13.1k21230

13.1k21230

asked Jan 25 at 4:50

John TophamJohn Topham

63

asked Jan 25 at 4:50

John TophamJohn Topham

63

asked Jan 25 at 4:50

John TophamJohn Topham

63

63

closed as off-topic by Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos Jan 25 at 13:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos Jan 25 at 13:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, what happens when you do complete the square?
  $endgroup$
  – Lord Shark the Unknown
  Jan 25 at 4:51
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, what happens when you do complete the square?
  $endgroup$
  – Lord Shark the Unknown
  Jan 25 at 4:51
  

  
  
  
  

$begingroup$
So, what happens when you do complete the square?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:51

$begingroup$
So, what happens when you do complete the square?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:51

add a comment | 

5 Answers
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If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:

$$x^2+ax+a=(x-frac a2)^2-frac{a^2}{4}+a$$

The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $ain(0,4)$, as stated in the question.

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answered Jan 25 at 5:05

Rhys HughesRhys Hughes

7,0401630

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is easiest for me to understand thanks
  $endgroup$
  – John Topham
  Jan 25 at 5:11
  

  
  
  
  

add a comment | 

0

$begingroup$

Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.

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answered Jan 25 at 4:54

Jimmy SabaterJimmy Sabater

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0

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Let $x^2+ax+a=y$

$iff x^2+ax+a-y=0$

As $x$ is real, the discriminant must be $ge0$

$a^2-4(a-y)ge0iff4yge a(4-a)$

It is sufficient to have $a(4-a)>0iff a(a-4)<0iff0<a<4$

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answered Jan 25 at 4:58

lab bhattacharjeelab bhattacharjee

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Complete the square:
$$x^2+ax+a>0 iff left(x+frac a2right)^2-frac{a^2}{4}+a>0 iff \
left(x+frac a2right)^2>frac{a^2-4a}{4} iff left(x+frac a2right)^2ge 0>frac{a^2-4a}{4}, 0<a<4.$$

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answered Jan 25 at 5:03

farruhotafarruhota

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0

$begingroup$

After completing the square of the above equation you will get:
$$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$
which is nothing but:
$$(x+frac{a}{2})^2 + (a-frac{a^2}{4})$$
The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+frac{a}{2})^2gt0$$
Now for the second part
$$a-frac{a^2}{4}gt0$$
which gives $$4a-a^2gt0$$
Or $$a^2-4alt0$$
$$a(a-4)lt0$$ which gives the critical points $a=0,4$
Now this inequality only holds if $0lt alt4$

Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers.
One more thing is you can check for the vertex of the parabola....
Hope this helps

share|cite|improve this answer

answered Jan 25 at 5:08

SNEHIL SANYALSNEHIL SANYAL

644110

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add a comment | 

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

$begingroup$

If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:

$$x^2+ax+a=(x-frac a2)^2-frac{a^2}{4}+a$$

The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $ain(0,4)$, as stated in the question.

share|cite|improve this answer

answered Jan 25 at 5:05

Rhys HughesRhys Hughes

7,0401630

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is easiest for me to understand thanks
  $endgroup$
  – John Topham
  Jan 25 at 5:11
  

  
  
  
  

add a comment | 

0

$begingroup$

If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:

$$x^2+ax+a=(x-frac a2)^2-frac{a^2}{4}+a$$

The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $ain(0,4)$, as stated in the question.

share|cite|improve this answer

answered Jan 25 at 5:05

Rhys HughesRhys Hughes

7,0401630

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is easiest for me to understand thanks
  $endgroup$
  – John Topham
  Jan 25 at 5:11
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:

$$x^2+ax+a=(x-frac a2)^2-frac{a^2}{4}+a$$

The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $ain(0,4)$, as stated in the question.

share|cite|improve this answer

answered Jan 25 at 5:05

Rhys HughesRhys Hughes

7,0401630

$endgroup$

If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:

$$x^2+ax+a=(x-frac a2)^2-frac{a^2}{4}+a$$

The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $ain(0,4)$, as stated in the question.

share|cite|improve this answer

answered Jan 25 at 5:05

Rhys HughesRhys Hughes

7,0401630

share|cite|improve this answer

share|cite|improve this answer

answered Jan 25 at 5:05

Rhys HughesRhys Hughes

7,0401630

answered Jan 25 at 5:05

Rhys HughesRhys Hughes

7,0401630

answered Jan 25 at 5:05

Rhys HughesRhys Hughes

7,0401630

7,0401630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is easiest for me to understand thanks
  $endgroup$
  – John Topham
  Jan 25 at 5:11
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is easiest for me to understand thanks
  $endgroup$
  – John Topham
  Jan 25 at 5:11
  

  
  
  
  

$begingroup$
This answer is easiest for me to understand thanks
$endgroup$
– John Topham
Jan 25 at 5:11

$begingroup$
This answer is easiest for me to understand thanks
$endgroup$
– John Topham
Jan 25 at 5:11

add a comment | 

0

$begingroup$

Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.

share|cite|improve this answer

answered Jan 25 at 4:54

Jimmy SabaterJimmy Sabater

3,023325

$endgroup$

add a comment | 

0

$begingroup$

Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.

share|cite|improve this answer

answered Jan 25 at 4:54

Jimmy SabaterJimmy Sabater

3,023325

$endgroup$

add a comment | 

0

0

0

$begingroup$

Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.

share|cite|improve this answer

answered Jan 25 at 4:54

Jimmy SabaterJimmy Sabater

3,023325

$endgroup$

Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.

share|cite|improve this answer

answered Jan 25 at 4:54

Jimmy SabaterJimmy Sabater

3,023325

share|cite|improve this answer

share|cite|improve this answer

answered Jan 25 at 4:54

Jimmy SabaterJimmy Sabater

3,023325

answered Jan 25 at 4:54

Jimmy SabaterJimmy Sabater

3,023325

answered Jan 25 at 4:54

Jimmy SabaterJimmy Sabater

3,023325

3,023325

add a comment | 

add a comment | 

0

$begingroup$

Let $x^2+ax+a=y$

$iff x^2+ax+a-y=0$

As $x$ is real, the discriminant must be $ge0$

$a^2-4(a-y)ge0iff4yge a(4-a)$

It is sufficient to have $a(4-a)>0iff a(a-4)<0iff0<a<4$

share|cite|improve this answer

answered Jan 25 at 4:58

lab bhattacharjeelab bhattacharjee

227k15158275

$endgroup$

add a comment | 

0

$begingroup$

Let $x^2+ax+a=y$

$iff x^2+ax+a-y=0$

As $x$ is real, the discriminant must be $ge0$

$a^2-4(a-y)ge0iff4yge a(4-a)$

It is sufficient to have $a(4-a)>0iff a(a-4)<0iff0<a<4$

share|cite|improve this answer

answered Jan 25 at 4:58

lab bhattacharjeelab bhattacharjee

227k15158275

$endgroup$

add a comment | 

0

0

0

$begingroup$

Let $x^2+ax+a=y$

$iff x^2+ax+a-y=0$

As $x$ is real, the discriminant must be $ge0$

$a^2-4(a-y)ge0iff4yge a(4-a)$

It is sufficient to have $a(4-a)>0iff a(a-4)<0iff0<a<4$

share|cite|improve this answer

answered Jan 25 at 4:58

lab bhattacharjeelab bhattacharjee

227k15158275

$endgroup$

Let $x^2+ax+a=y$

$iff x^2+ax+a-y=0$

As $x$ is real, the discriminant must be $ge0$

$a^2-4(a-y)ge0iff4yge a(4-a)$

It is sufficient to have $a(4-a)>0iff a(a-4)<0iff0<a<4$

share|cite|improve this answer

answered Jan 25 at 4:58

lab bhattacharjeelab bhattacharjee

227k15158275

share|cite|improve this answer

share|cite|improve this answer

answered Jan 25 at 4:58

lab bhattacharjeelab bhattacharjee

227k15158275

answered Jan 25 at 4:58

lab bhattacharjeelab bhattacharjee

227k15158275

answered Jan 25 at 4:58

lab bhattacharjeelab bhattacharjee

227k15158275

227k15158275

add a comment | 

add a comment | 

0

$begingroup$

Complete the square:
$$x^2+ax+a>0 iff left(x+frac a2right)^2-frac{a^2}{4}+a>0 iff \
left(x+frac a2right)^2>frac{a^2-4a}{4} iff left(x+frac a2right)^2ge 0>frac{a^2-4a}{4}, 0<a<4.$$

share|cite|improve this answer

answered Jan 25 at 5:03

farruhotafarruhota

21.2k2841

$endgroup$

add a comment | 

0

$begingroup$

Complete the square:
$$x^2+ax+a>0 iff left(x+frac a2right)^2-frac{a^2}{4}+a>0 iff \
left(x+frac a2right)^2>frac{a^2-4a}{4} iff left(x+frac a2right)^2ge 0>frac{a^2-4a}{4}, 0<a<4.$$

share|cite|improve this answer

answered Jan 25 at 5:03

farruhotafarruhota

21.2k2841

$endgroup$

add a comment | 

0

0

0

$begingroup$

Complete the square:
$$x^2+ax+a>0 iff left(x+frac a2right)^2-frac{a^2}{4}+a>0 iff \
left(x+frac a2right)^2>frac{a^2-4a}{4} iff left(x+frac a2right)^2ge 0>frac{a^2-4a}{4}, 0<a<4.$$

share|cite|improve this answer

answered Jan 25 at 5:03

farruhotafarruhota

21.2k2841

$endgroup$

Complete the square:
$$x^2+ax+a>0 iff left(x+frac a2right)^2-frac{a^2}{4}+a>0 iff \
left(x+frac a2right)^2>frac{a^2-4a}{4} iff left(x+frac a2right)^2ge 0>frac{a^2-4a}{4}, 0<a<4.$$

share|cite|improve this answer

answered Jan 25 at 5:03

farruhotafarruhota

21.2k2841

share|cite|improve this answer

share|cite|improve this answer

answered Jan 25 at 5:03

farruhotafarruhota

21.2k2841

answered Jan 25 at 5:03

farruhotafarruhota

21.2k2841

answered Jan 25 at 5:03

farruhotafarruhota

21.2k2841

21.2k2841

add a comment | 

add a comment | 

0

$begingroup$

After completing the square of the above equation you will get:
$$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$
which is nothing but:
$$(x+frac{a}{2})^2 + (a-frac{a^2}{4})$$
The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+frac{a}{2})^2gt0$$
Now for the second part
$$a-frac{a^2}{4}gt0$$
which gives $$4a-a^2gt0$$
Or $$a^2-4alt0$$
$$a(a-4)lt0$$ which gives the critical points $a=0,4$
Now this inequality only holds if $0lt alt4$

Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers.
One more thing is you can check for the vertex of the parabola....
Hope this helps

share|cite|improve this answer

answered Jan 25 at 5:08

SNEHIL SANYALSNEHIL SANYAL

644110

$endgroup$

add a comment | 

0

$begingroup$

After completing the square of the above equation you will get:
$$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$
which is nothing but:
$$(x+frac{a}{2})^2 + (a-frac{a^2}{4})$$
The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+frac{a}{2})^2gt0$$
Now for the second part
$$a-frac{a^2}{4}gt0$$
which gives $$4a-a^2gt0$$
Or $$a^2-4alt0$$
$$a(a-4)lt0$$ which gives the critical points $a=0,4$
Now this inequality only holds if $0lt alt4$

Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers.
One more thing is you can check for the vertex of the parabola....
Hope this helps

share|cite|improve this answer

answered Jan 25 at 5:08

SNEHIL SANYALSNEHIL SANYAL

644110

$endgroup$

add a comment | 

0

0

0

$begingroup$

After completing the square of the above equation you will get:
$$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$
which is nothing but:
$$(x+frac{a}{2})^2 + (a-frac{a^2}{4})$$
The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+frac{a}{2})^2gt0$$
Now for the second part
$$a-frac{a^2}{4}gt0$$
which gives $$4a-a^2gt0$$
Or $$a^2-4alt0$$
$$a(a-4)lt0$$ which gives the critical points $a=0,4$
Now this inequality only holds if $0lt alt4$

Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers.
One more thing is you can check for the vertex of the parabola....
Hope this helps

share|cite|improve this answer

answered Jan 25 at 5:08

SNEHIL SANYALSNEHIL SANYAL

644110

$endgroup$

After completing the square of the above equation you will get:
$$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$
which is nothing but:
$$(x+frac{a}{2})^2 + (a-frac{a^2}{4})$$
The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+frac{a}{2})^2gt0$$
Now for the second part
$$a-frac{a^2}{4}gt0$$
which gives $$4a-a^2gt0$$
Or $$a^2-4alt0$$
$$a(a-4)lt0$$ which gives the critical points $a=0,4$
Now this inequality only holds if $0lt alt4$

Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers.
One more thing is you can check for the vertex of the parabola....
Hope this helps

share|cite|improve this answer

answered Jan 25 at 5:08

SNEHIL SANYALSNEHIL SANYAL

644110

share|cite|improve this answer

share|cite|improve this answer

answered Jan 25 at 5:08

SNEHIL SANYALSNEHIL SANYAL

644110

answered Jan 25 at 5:08

SNEHIL SANYALSNEHIL SANYAL

644110

answered Jan 25 at 5:08

SNEHIL SANYALSNEHIL SANYAL

644110

644110

add a comment | 

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