# Math Help - quadratics and constant K

1. ## quadratics and constant K

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?

2. Hi.

Originally Posted by furor celtica
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?
I guess you know how to sketch y = x-1.
What about the quadratic function k(x^2) ? I guess you know it has a maximum or minimum in x = 0.

if k < 0 the parabola opens downward. So the curve kx^2 has a maximum in x=0.

If you substitute x = 0 in y = x-1 it is 0-1 = -1, so the line is below the quadratic function and there got to be two solutions, because y = x-1 is a linear function.

Otherwise you can show it directly:

$x - 1 = kx^2$

$-kx^2 +x-1 = 0$

$kx^2 - x +1 = 0$

$x^2 - \frac{x}{k} + \frac{1}{k} = 0$

$x_{1,2} = + \frac{1}{2k} \pm \sqrt{(\frac{1}{2k})^2-\frac{1}{k}}$

Assume k < 0. There are two solutions, if

$\sqrt{(\frac{1}{2k})^2-\frac{1}{k}} > 0$

$\left(\sqrt{(\frac{1}{2k})^2-\frac{1}{k}} \right)^2 > 0^2$

$\frac{1}{4k^2})-\frac{1}{k}$

Of course this is > 0, because let k:= -1

then

$k^2 = (-1)^2 = 1$

and $- \frac{1}{k} = - \frac{1}{-1} =1$

If you substitute -a = k (if k=-1 the a is equal to -(-1) = +1), you get

$\frac{1}{4(-a)^2} + \frac{1}{a} = \frac{1}{4a^2} + \frac{1}{a}$

This could be kinda confusing...

Yours
Rapha

3. but what about the root between 0 and 1? i dont understand that

4. Solving the quadratic equation is one way to go.

But consider that the line y=x-1 goes from (0,-1) to (1,0) and is continuous and increasing on this interval.

And consider that y=kx^2 goes from (0,0) to (1,k) and is continuous and decreasing on this interval, and k < 0.

So the curves must cross (by hand-wavy "can't lift your pencil" argument... since the problem statement allows graphical proof, this should be enough).

Similar for the other root that occurs where x < 0. The curve y=kx^2 gets "very steep" as x approaches negative infinity, so it has to cross the line at some point.

5. i'd like to understand how to do this using the quadratic formula, but unfortunately rapha's explanation wasn't very clear to me
for example, why did you divide by k before starting? can't you just end up with (1+-sqrt(1-4k))/2k? i got to that but i dont really know how it fits in

6. Originally Posted by furor celtica
i'd like to understand how to do this using the quadratic formula, but unfortunately rapha's explanation wasn't very clear to me
for example, why did you divide by k before starting? can't you just end up with (1+-sqrt(1-4k))/2k? i got to that but i dont really know how it fits in
No offense to Rapha intended, but I also didn't see the motivation behind some of Rapha's methodology.

Here's how I would proceed.

$x-1=kx^2$

$kx^2-x+1=0$

$x=\frac{1\pm \sqrt{1-4k}}{2k}$

as you wrote.

Now we know immediately that there are two real solutions because $\sqrt{1-4k}$ is positive. It only remains to show that one of these is in the interval (0,1).

What we need to do is show that $\sqrt{1-4k}$ is greater than 1 and less than 1-2k. (To see why, plug in those values and see that we get x=0 and x=1 respectively. We are dealing with taking the minus where there is a "plus/minus" since taking the plus would result in x < 0.)

Clearly $\sqrt{1-4k}$ is greater than 1, because 1-4k is greater than one. Now consider that $(1-2k)^2 = 1-4k+4k^2 > 1-4k$. Therefore $\sqrt{1-4k} < 1-2k$ and we are done.