Discriminant Issue

[mrhmaths]

Hello all,

See attached photo for my problem. I would be very grateful for any help.



My understanding is that I can divide across by the negative 4 in step 4 above out of convention to make things a little easier to work with but I can't get my head around why this is affecting my answer.

Am I missing some sort of fundamental transformation that's occurring to the quadratic?

 

[Deleted member 4993]

Hello all,

See attached photo for my problem. I would be very grateful for any help.

View attachment 34152

My understanding is that I can divide across by the negative 4 in step 4 above out of convention to make things a little easier to work with but I can't get my head around why this is affecting my answer.

Am I missing some sort of fundamental transformation that's occurring to the quadratic?

When you divide by a negative number, the inequality sign flips. Then we have,

-8p² - 4q² > 0 ........... divide both sides by -4 and flip the inequality sign to get,

2p² + q² < 0 ....... no real soution.

 

[Dr.Peterson]

I think you're just not paying attention to the meaning of what you did.

Your two inequalities, [imath]-2p^2-q^2<0[/imath] and [imath]2p^2+q^2>0[/imath], are equivalent. Both are true for all real p and q, so both tell you that the discriminant is negative for all real p and q.

I think you're forgetting that the LHS of these inequalities is not the discriminant itself! What;'s important is not whether it is positive or negative, but whether the entire inequality is true or false.

There is also a little error in the way you stated this. I would change the start to

​

That is, you don't want to show that IF there are no real roots, THEN such and such; you want to show something to be true which will IMPLY that there are no real roots. So the goal is a series of equivalent statements: there are no real roots if and only if the discriminant is negative, which is true if and only if ... an inequality that is always true.

 

[Harry_the_cat]

You are trying to show that the discriminant < 0. You don't know that it is <0 so your second line

is not correct at this stage because you don't know that the discriminant <0, you are trying to show that.

What you do know is that the discriminant is equal to
.
Now, you can say that that

\(\displaystyle -8p^2-4q^2 = -4(2p^2+q^2) \le0\) because the bracketed bit is always \(\displaystyle \ge0\) for real p and q.

[Note that if p and q are both 0, the equation becomes \(\displaystyle x^2=0\) which does have a real root. This is the one exception to the statement, so it is NOT TRUE in general.]

The problem with your "proof", besides assuming that what you are trying to prove is true in your second line, is that when you divide by \(\displaystyle -4\), the expression on your left is no longer equal to the discriminant. In fact, the same applies when you divide by \(\displaystyle +4\).

 

[Harry_the_cat]

When you divide by a negative number, the inequality sign flips. Then we have,

-8p² - 4q² > 0 ........... divide both sides by -4 and flip the inequality sign to get,

2p² + q² < 0 ....... no real soution.

But mrhmaths has correctly flipped the sign when dividing by -4. I think you may have misread the signs.

Your two inequalities, -2p^2-q^2<0−2p2−q2<0 and 2p^2+q^2>02p2+q2>0, are equivalent. Both are true for all real p and q, so both tell you that the discriminant is negative for all real p and q.

But these statements are NOT true if p=q=0.

 

[Dr.Peterson]

But these statements are NOT true if p=q=0.

Excellent point, which I missed, probably assuming the OP stated the assignment correctly, and therefore neglecting to check whether it was true.

So the equation x^2 = 0 does have a real root, and the claim is false.

 

[Steven G]

You do NOT know if the discriminant is negative or not. That is why you need to put a question mark, ?, over each inequality until you know if one is in fact true!

 

[Dr.Peterson]

You do NOT know if the discriminant is negative or not. That is why you need to put a question mark, ?, over each inequality until you know if one is in fact true!

That's what I do when I check an answer. We don't know that the answer is correct until we finish evaluating.

Here, we are trying to prove something, so I would prefer what I suggested, which is to clearly indicate the direction of implication, which needs to be from the final inequality back to the original claim (or, when valid, that they are all equivalent):

So the goal is a series of equivalent statements: there are no real roots if and only if the discriminant is negative, which is true if and only if ... an inequality that is always true.

In other cases, you might have to say, "My claim is true if this is true, which is true if that is true, ..."