discriminant (1 Viewer)

[Aesytic]

for what values of p does the equation px^2 - 4x + p = 0 have real roots?
the answer is -2<=p<=2, but i was wondering, can p=0? because it just asks for real roots, and if p=0 then x=0, but does that still count?

 

[marksgm]

yes p can equal 0!

 

[marksgm]

since the equation is quadratic AND has TWO real roots since discriminant >0 . working out through the quadratic formula, we get that the roots are 0 and 4p. Meaning the parabola px^2 - 4x + p^2 = 0 cuts the x-axis at (0,0) and (4p,0). so yes x can equal zero since the parabola passes through (0,0)

 

[Aesytic]

ah, ok, thanks for that!

edit: sorry, the constant is actually p. does that change anything?

 

[Trebla]

for what values of p does the equation px^2 - 4x + p = 0 have real roots?
the answer is -2<=p<=2, but i was wondering, can p=0? because it just asks for real roots, and if p=0 then x=0, but does that still count?

If p = 0 the equation is no longer a quadratic and therefore the usual idea of discriminants don't apply. This can be seen from the quadratic formula that



Note that p is non-zero for a solution to exist for the quadratic. So technically speaking when we look at the quadratic formula, for real roots we have



Now we consider the case when p = 0 in isolation (because the equation is no longer quadratic the above treatment is not appropriate). This implies x = 0 which is clearly a real root thus we include it in our domain which leads to

 

[cutemouse]

If p = 0 the equation is no longer a quadratic and therefore the usual idea of discriminants don't apply.

Well, if one has done a bit of linear algebra, then one would conclude that it is a polynomial in P^2 (R)... ie a vector (s,t,0)^T with respect to the standard basis {1,x,x^2} for s,t real...

So it still could be a quadratic. It just depends how you define one...

 

[Aesytic]

If p = 0 the equation is no longer a quadratic and therefore the usual idea of discriminants don't apply. This can be seen from the quadratic formula that



Note that p is non-zero for a solution to exist for the quadratic. So technically speaking when we look at the quadratic formula, for real roots we have



Now we consider the case when p = 0 in isolation (because the equation is no longer quadratic the above treatment is not appropriate). This implies x = 0 which is clearly a real root thus we include it in our domain which leads to

yeah, i just checked with my teacher today, and he said the same thing. thanks for the help!