more on quadratics

**puresoul** · Dec 2nd 2012, 08:37 PM

a) the diagram shows the graphs of y=x-1 and y=kx^2 where k is a positive constant.
The graphs intersect at 2 dinstinct points A and B.
Write down the quadratic equation satisfied by the x-coordinates of A and B, and hence show that k<1/4

b) Describe briefly the relashioship between the graphs of y=x-1 and y=kx^2 in each of the cases
i) k=1/4
ii) k<1/4
c) Show, by using a graphical argument of otherwise that when k is a negative constant, the equation x-1=kx^2 has 2 real roots, one of which lies between 0 and 1.

I need help in the last part of the Q.

I know that to show that the equation has 2 real roots we need to show that its determinant is greater than zero ,but how do we know that one of those points are between 0 and 1??

**Prove It** · Dec 2nd 2012, 08:40 PM

Couldn't you simply plug into the Quadratic Formula?

**puresoul** · Dec 2nd 2012, 08:44 PM

Originally Posted by Prove It

Couldn't you simply plug into the Quadratic Formula?

You mean to find the 2 points??
But how do we then know if one of them is between 0 and 1
We don't know the value of k ?

I tried to plug it in and got x= (1+or- squreroot k^2 +4k )/ 2k

Is that what you mean?? Or do you mean something else?

**Prove It** · Dec 2nd 2012, 09:11 PM

Well your use of the Quadratic formula is incorrect. If $\displaystyle \begin{align*} k\,x^2 = x - 1 \end{align*}$ then $\displaystyle \begin{align*} k\,x^2 - x + 1 = 0 \end{align*}$ . Solving for x gives $\displaystyle \begin{align*} x= \frac{1 \pm \sqrt{1^2 - 4(k)(1)}}{2k} = \frac{1 \pm \sqrt{1 - 4k}}{2k} \end{align*}$

Now if $\displaystyle \begin{align*} k < 0 \end{align*}$ that means $\displaystyle \begin{align*} 1 - 4k > 0 \end{align*}$ , which means $\displaystyle \begin{align*} \sqrt{1 - 4k} > 0 \end{align*}$ .

From this we can state that $\displaystyle \begin{align*} \frac{1 + \sqrt{1 - 4k}}{2k} < 0 \end{align*}$ due to $\displaystyle \begin{align*} k < 0 \end{align*}$ . So that means it must be $\displaystyle \begin{align*} \frac{1 - \sqrt{1 - 4k}}{2k} \end{align*}$ that's between 0 and 1.

It should be obvious to start with that this fraction must be positive. If not, convince yourself of it.

Now let's assume that this was $\displaystyle \begin{align*} > 1 \end{align*}$ . Then we would have

$\displaystyle \begin{align*} \frac{1 - \sqrt{1 - 4k}}{2k} &> 1 \\ 1 - \sqrt{1 - 4k} &< 2k \\ 1 - 2k &< \sqrt{1 - 4k} \\ \left( 1 - 2k \right)^2 &< 1 - 4k \\ 1 - 4k + 4k^2 &< 1 - 4k \\ 4k^2 &< 0 \end{align*}$

This is an obvious contradiction, which means that $\displaystyle \begin{align*} \frac{1 - \sqrt{1 - 4k}}{2k} \end{align*}$ is NOT $\displaystyle \begin{align*} \geq 1 \end{align*}$ . Therefore it must be between 0 and 1.

**puresoul** · Dec 2nd 2012, 09:25 PM

I GOT It at last !!!
^__^

THANKS ALOT !!!

**MarkFL** · Dec 2nd 2012, 09:28 PM

When dealing with quadratic equations, we analyze the discriminant to find whether we have two distinct real roots, one real repeated root, or complex conjugate roots.

If the discriminant is positive, then we have two distinct real roots. The discriminant is

$D=1-4k$ and with $k<0$ we know:

$D>0$

The quadratic formula tells us the roots will be:

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

Let's examine the root:

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

$0<\frac{1-\sqrt{1-4k}}{2k}<1$

Multiply through by $0<-2k$

$0<\sqrt{1-4k}-1<-2k$

$1<\sqrt{1-4k}<1-2k$

$1<1-4k<1-4k+4k^2$

$0<-4k<-4k+4k^2$

$4k<0<4k^2$

$k<0<k^2$

Since this must be true, then the root we began with must be on the required interval.

**puresoul** · Dec 2nd 2012, 09:43 PM

Originally Posted by MarkFL2

When dealing with quadratic equations, we analyze the discriminant to find whether we have two distinct real roots, one real repeated root, or complex conjugate roots.

If the discriminant is positive, then we have two distinct real roots. The discriminant is

$D=1-4k$ and with $k<0$ we know:

$D>0$

The quadratic formula tells us the roots will be:

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

Let's examine the root:

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

$0<\frac{1-\sqrt{1-4k}}{2k}<1$

Multiply through by $0<-2k$

$0<\sqrt{1-4k}-1<-2k$

$1<\sqrt{1-4k}<1-2k$

$1<1-4k<1-4k+4k^2$

$0<-4k<-4k+4k^2$

$4k<0<4k^2$

$k<0<k^2$

Since this must be true, then the root we began with must be on the required interval.

Thnks
I understood what you did ,
although I didn't knowbefore that it was possible to solve an inequality without separating it into 2 inequalities and solving each one separately :O

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y=x-1 and y=kx^2 show that k <1/4

the diagram shows the graphs of y=x-1 and y=kx^2 where k is a positive constant. The graphs intersect at 2 dinstinct points A and B.

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