Quadratic Proof: Showing a>0 Implies b^2 - 4ac \leq 0

[zooxanthellae]

Quadratic "Proof"

Homework Statement

Show that if $$a>0$$, then $$ax^2 + bx + c \geq 0$$ for all values of $$x$$ if and only if $$b^2 - 4ac \leq 0$$

http://www.math.toronto.edu/~drorbn/classes/0405/157AnalysisI/HW2/HW.html

Homework Equations

I believe I'm supposed to be working only with basic rules like the commutative and distributive laws.

The Attempt at a Solution

I re-arranged $$b^2 - 4ac \leq 0$$ to $$b^2 \leq 4ac$$.

Then I took the square root of both sides to get $$b \leq 2\sqrt{ac}$$.

I thought about this result and concluded that this meant that $$b/2$$ is less than some number between $$a$$ and $$c$$. But I couldn't see where to go from there.

Then I tried working with the equation I was given in order to see if I could manipulate it so that $$b^2 - 4ac \leq 0$$ would have to be true.

So I re-arranged $$ax^2 + bx + c \geq 0$$ to $$ax^2 + c \geq |-bx|$$. And then I found myself stuck once again.

I think my problem is with the $$x^2$$ and $$x$$. I don't see how the preconditions given really "address" the difference between the two.

 

[Tedjn]

How can you use the quadratic formula to help?

 

[Mark44]

Homework Statement

Show that if $$a>0$$, then $$ax^2 + bx + c \geq 0$$ for all values of $$x$$ if and only if $$b^2 - 4ac \leq 0$$

http://www.math.toronto.edu/~drorbn/classes/0405/157AnalysisI/HW2/HW.html

Homework Equations

I believe I'm supposed to be working only with basic rules like the commutative and distributive laws.

This might not be a valid assumption.

The Attempt at a Solution

I re-arranged $$b^2 - 4ac \leq 0$$ to $$b^2 \leq 4ac$$.

Then I took the square root of both sides to get $$b \leq 2\sqrt{ac}$$.

This is not the whole story. If b² <= 4ac, then
$$-2\sqrt{ac} \leq b \leq 2\sqrt{ac}$$

I thought about this result and concluded that this meant that $$b/2$$ is less than some number between $$a$$ and $$c$$. But I couldn't see where to go from there.

Then I tried working with the equation I was given in order to see if I could manipulate it so that $$b^2 - 4ac \leq 0$$ would have to be true.

So I re-arranged $$ax^2 + bx + c \geq 0$$ to $$ax^2 + c \geq |-bx|$$. And then I found myself stuck once again.

I think my problem is with the $$x^2$$ and $$x$$. I don't see how the preconditions given really "address" the difference between the two.

This is an if and only if proof, meaning that there are two parts you need to prove. Tedjn's advice to think about the quadratic formula is good, unless there is some specific reason you aren't supposed to use it.

 

[zooxanthellae]

How can you use the quadratic formula to help?

As far as I can tell I'm not supposed to, but assuming that I am:

If we suppose that $$b^2 - 4ac \leq 0$$ than that invalidates the $$\sqrt{b^2 - 4ac}$$ portion of the quadratic formula. Couldn't this mean that $$ax^2 + bx + c > 0$$, since the quadratic formula requires that the equation = 0?

EDIT: The reason I'm hesitant about the quadratic formula is that this problem comes from a homework assignment that deals with Chapter 1 of Spivak's Calculus. The assignment specifies 3 problems from Spivak and 3 problems from the class, but none of the 3 Spivak problems are like this, and this problem is the first of the 3.

 

[Tedjn]

I don't think it's a problem using the quadratic formula. Note that you can derive it by completing the square, which doesn't use anything at all complicated.

Let's take this one step at a time. Your intuition is almost correct. It might help to consider two cases, when b² - 4ac < 0 and when b² - 4ac = 0. In the first case, the quadratic can never equal 0. (Why?) Since a > 0, the quadratic will be positive in at least one point. (Where and why?) Quadratic equations are continuous. Can you use this to show that the function is positive everywhere?

 

[HallsofIvy]

By completing the square (which requires only basic arithmetic operations), [itex]ax^2+ bx+ c= a(x^2+ (b/a)x+ (b^2/4a^2)- (b^2/4a^2))+ c= a(x^2+ b/2a)+ (c- b^/4a). When is that the sum of two positive numbers?

 

[D H]

Per post #4, this is a calculus problem. That suggests using calculus. What is the value of the quadratic where the derivative is zero, and what is the meaning of this value?

 

[Tedjn]

@HallsofIvy and D H: Thanks for pointing this out -- a much better method.

 

[stronghold.mr]

I HOPE YOU ARE CLEAR WITH THE GRAPHS OF PARABOLA FOR QUADRATIC EQUATION....IF YOU ARE NOT SEE THE LINK BELOW(NOT VERY DESCRIPTIVE BUT THEN ALSO HELPFUL

SEE THIS http://picasaweb.google.com/lh/photo/2LXaGSP0A9V_AI5_vVAOcQ?feat=directlink" (NOT ABLE TO ATTACH IT) -
http://picasaweb.google.com/lh/photo/2LXaGSP0A9V_AI5_vVAOcQ?feat=directlink

 

[stronghold.mr]

AFTER SEEING TO THE LINKS NOW WE MAY PROCEED

AS WE KNOW THAT FOR PARABOLA
ABOVE X AXIS ALL THE VALUES OF function y=ax^2+bx+c > 0

BELOW X AXIS ALL THE VALUES OF function y=ax^2+bx+c < 0





now i hope you can join every knowledge you gained and get your answer

reply if want something more

 

[Mark44]

I HOPE YOU ARE CLEAR WITH THE GRAPHS OF PARABOLA FOR QUADRATIC EQUATION....IF YOU ARE NOT SEE THE LINK BELOW(NOT VERY DESCRIPTIVE BUT THEN ALSO HELPFUL

Writing in all caps is considered "shouting." It's even worse when you use a large font size.

SEE THIS http://picasaweb.google.com/lh/photo/2LXaGSP0A9V_AI5_vVAOcQ?feat=directlink" (NOT ABLE TO ATTACH IT) -
http://picasaweb.google.com/lh/photo/2LXaGSP0A9V_AI5_vVAOcQ?feat=directlink

The drawings at the linked site are only marginally helpful, IMO.

 

[stronghold.mr]

Writing in all caps is considered "shouting." It's even worse when you use a large font size.

The drawings at the linked site are only marginally helpful, IMO.

sorry mark

didnt know that

will take care from next time

just wanted to make a bit attractive

 

[zooxanthellae]

I don't think it's a problem using the quadratic formula. Note that you can derive it by completing the square, which doesn't use anything at all complicated.

Let's take this one step at a time. Your intuition is almost correct. It might help to consider two cases, when b² - 4ac < 0 and when b² - 4ac = 0. In the first case, the quadratic can never equal 0. (Why?) Since a > 0, the quadratic will be positive in at least one point. (Where and why?) Quadratic equations are continuous. Can you use this to show that the function is positive everywhere?

I've worked at it for a while and still can't seem to even show that $$ax^2 + bx + c$$ is never zero when $$b^2 - 4ac < 0$$. That assumption leads me to say that $$4ac > b^2$$ and then that $$-2\sqrt{ac} < b < 2\sqrt{ac}$$. Therefore since we assumed that $$4ac > b^2$$ and $$a$$ is positive, $$c$$ must be positive as well. Therefore when $$x=0$$ the whole equation would be positive due to $$c$$. So if we're trying to prove that the quadratic is never 0, and we know that the quadratic has a positive point, then we must prove that the quadratic is always positive, correct? Then if the quadratic is always positive, $$ax^2 + c > |bx|$$, or $$ax^2 + c > |2\sqrt{ac}x|$$. And I'm stuck again.

 

[zooxanthellae]

By completing the square (which requires only basic arithmetic operations), ax^2+ bx+ c= a(x^2+ (b/a)x+ (b^2/4a^2)- (b^2/4a^2))+ c= a(x^2+ b/2a)+ (c- b^/4a). When is that the sum of two positive numbers?

Can you explain (sorry) how you got from $$a(x^2 + (b/a)x + (b^2/4a^2) - (b^2/4a^2)) + c$$ to $$a(x^2 + b/2a) + (c - b^2/4a)$$? The second one equals $$ax^2 + b/2 + c - b^2/4a$$. How is that the same as the first one, which was $$ax^2 + bx + c$$?

 

[zooxanthellae]

Per post #4, this is a calculus problem. That suggests using calculus. What is the value of the quadratic where the derivative is zero, and what is the meaning of this value?

You are correct that the problem comes from a course that uses a Calculus textbook. However, the problem was assigned in the very early portions of the course, when limits haven't even been mentioned. Therefore I don't think I'm supposed to use Calculus.

If you like, here is the course I am using:

http://www.math.toronto.edu/~drorbn/classes/0405/157AnalysisI/index.html

Note that I am on Homework Assignment 2, whereas Limits are first mentioned around Homework Assignment 5.

 

[Mark44]

I've worked at it for a while and still can't seem to even show that $$ax^2 + bx + c$$ is never zero when $$b^2 - 4ac < 0$$.

If b² - 4ac < 0, there are no real roots to the quadratic equation, so there are no (real) x intercepts. This means that the graph of the quadratic lies completely above or completely below the x-axis. If a > 0, the graph is above the x-axis, meaning that ax² + bx + c > 0 for all x. If a < 0, the graph is below the x-axis, meaning that ax² + bx + c < 0 for all x.

Below, it makes no sense to say that "the whole equation would be positive..." An equation is neither positive, negative, nor zero. It is either true or false.

That assumption leads me to say that $$4ac > b^2$$ and then that $$-2\sqrt{ac} < b < 2\sqrt{ac}$$. Therefore since we assumed that $$4ac > b^2$$ and $$a$$ is positive, $$c$$ must be positive as well. Therefore when $$x=0$$ the whole equation would be positive due to $$c$$. So if we're trying to prove that the quadratic is never 0, and we know that the quadratic has a positive point, then we must prove that the quadratic is always positive, correct? Then if the quadratic is always positive, $$ax^2 + c > |bx|$$, or $$ax^2 + c > |2\sqrt{ac}x|$$. And I'm stuck again.

 

[zooxanthellae]

If b² - 4ac < 0, there are no real roots to the quadratic equation, so there are no (real) x intercepts. This means that the graph of the quadratic lies completely above or completely below the x-axis. If a > 0, the graph is above the x-axis, meaning that ax² + bx + c > 0 for all x. If a < 0, the graph is below the x-axis, meaning that ax² + bx + c < 0 for all x.

Below, it makes no sense to say that "the whole equation would be positive..." An equation is neither positive, negative, nor zero. It is either true or false.

Ahhh finally was able to derive the quadratic formula and now I think this makes sense.

Sorry about the misstatement about the equation.

Thanks for the help!