quadratics and constant K

[red and white kop!]

given the line y=x-1 and the curve y=k(x^2)
show, by using a graphical argument or otherwise, that when k is a negative constant the equation k(x^2)=x-1 has two real roots, one of which lies between 0 and 1

this question followed a series of straightforward quadratics problems but here i'm a bit lost, i don't even know how to work it through
can someone show me?

 

[Deleted member 4993]

given the line y=x-1 and the curve y=k(x^2)
show, by using a graphical argument or otherwise, that when k is a negative constant the equation k(x^2)=x-1 has two real roots, one of which lies between 0 and 1

this question followed a series of straightforward quadratics problems but here i'm a bit lost, i don't even know how to work it through
can someone show me?

let k = -m (where 'm' is a positive constant)

mx[sup:z121pqe5]2[/sup:z121pqe5] + x - 1 = 0

\(\displaystyle x_{1,2} \ \ = \ \ \frac{-(1) \pm \sqrt{(1)^2 - 4(-1)(m)}}{2m}\)

continue...

 

[BigGlenntheHeavy]

\(\displaystyle Good \ problem, \ red \ and \ white \ kop!.\)

\(\displaystyle y \ = \ kx^2 \ and \ y \ = \ x-1 \ \implies \ \ \ kx^2-x+1 \ = \ 0\)

\(\displaystyle Hence, \ x \ = \ \frac{1\pm\sqrt{1-4k}}{2k}\)

\(\displaystyle Now, \ to \ have \ two \ distinct \ real \ roots, \ the \ discriminate \ must \ be \ greater \ than \ zero.\)

\(\displaystyle Ergo, \ 1-4k \ > \ 0 \ \implies \ 4k \ < \ 1, \ k \ < \ \frac{1}{4}, \ k \ \ne 0, \ (division \ by \ 0 \ a \ no-no)\)

\(\displaystyle However, \ no \ problem, \ as \ we \ let \ k \ < \ 0 \ as \ prescribed.\)

\(\displaystyle Hence, \ \lim_{k\to-\infty} \frac{1-\sqrt{1-4k}}{2k} \ = \ 0 \ and \ \lim_{k\to0^-} \frac{1-\sqrt{1-4k}}{2k} \ = \ 1\)

\(\displaystyle Hence, \ one \ of \ the \ solutions \ will \ always \ be \ between \ 0 \ and \ 1. \ QED\)

\(\displaystyle Range \ = \ (0,1) \ and \ is \ smooth \ with \ no \ critical \ points, \ see \ graph \ below.\)

[attachment=0:1mfkvggb]iii.jpg[/attachment:1mfkvggb]