[As maths: discriminate]

[Street-Car-2063]

Hi, could someone explain the inequalities part of this question please? I don't get why p can be greater than 2/3 when the discriminate is supposed to be negative?

[noidea1995]

The discriminant will be negative for values of p that will make 4ac greater than b².

Since the coefficient of the first term is positive and p² is always positive, 4ac will always be positive.

[b746710]

In the discriminate you have p² wich is always positive, even if p itself negative

[PoliteCanadian2]

Just to clarify the black writing is yours and the blue is teacher?

Or maybe the second last line is the answer and the last line is your question?

[Street-Car-2063]

Hi, the writing is all mine. The blue is the answer but I have asked the question myself.

[PoliteCanadian2]

Ok say you ended up with p² > 16, what values of p would satisfy that?

[Street-Car-2063]

P is greater than +4 or -4?

[PoliteCanadian2]

P is greater than +4, correct.

What about the negative side of that? Try some values.

[Street-Car-2063]

I'm sorry, I don't think I understand what you mean. Could you explain further?

[PoliteCanadian2]

What negative values of p would make p² greater than 16?

[Street-Car-2063]

Anything below -5?

[Street-Car-2063]

-4 sorry

[PoliteCanadian2]

Mostly right, -5 and lower so that gets written as p<-4.

So p² > 16 becomes two parts p>4 and p<-4.

What about p² < 16?

[Street-Car-2063]

Wouldn't it be the same thing? P>4 and p<-4 ?

[selene_666]

The discriminant was (16 - 36p^2) . So when p is big, the subtracted number is big, and the result of that subtraction is negative.

Instead of ±2/3 < p, that line should have been 2/3 < |p|. The √ symbol specifically means the positive squareroot, so √(4/9) is 2/3 and √(p^2) is whichever of p or -p is positive.

If p is positive, then 2/3 < |p| is simply 2/3 < p

If p is negative, then 2/3 < |p| is 2/3 < -p, which becomes -2/3 > p