polynomial q (1 Viewer)

[Patato]

this may be obvious to some..but im not seeing it

x^2 - x + k = 0

it has two distinct positive real roots

question is to show 0 < k < 1/4

i can do this by showing the discriminant is > 0 to show it has two distinct positive real roots...but why is it greater than zero?

is it because if it were < than zero, it would mean the sum of the roots two at a time would be negative, which isnt possible since both roots are positive?

 

[ohexploitable]

^
Quadratic formula, x only has distinct real values if the discriminant is greater than zero.

 

[Patato]

please read my whole post lol, i already got that much, but the question is asking to show 0 < k < 1/4

thx anyway

 

[ohexploitable]

oh wait... my bad...

 

[ohexploitable]

Okay, from the quadratic formula,



-----

For the '+' case,



This is always positive so we don't have to worry about it.

-----

For the '-' case,



The root is only positive if,



Substitute in the discriminant and you should find k>0.

 

[khorne]

Let

y-k = x(x-1)

So k = 0 y = x(x-1) has roots 0 and 1

As k increases, the parabola is shifted upwards, k units. If k is 1/4, it is a perfect square (x-1/2)^2, with one positive solution 1/2, so therefore, 0< k <1/4

 

[ZachBC_94]

Ok, so x^2 - x + k = 0

Therefore:

delta = 1^2 - 4k

= 1 - 4k

Now, for one root, delta = 0

For no roots, delta < 0

If k >= 1/4, delta <= 0 (a.k.a. 1 or 0 roots)

Note if k < 0, one root will be always be negative

So if you want to count 0 as not positive, then 0 < k < 1/4