use the discriminant to find the range of values that K can take if the equation K$\displaystyle x^2$-4x+5-K=0 has no real roots

:eek: someone please help!!! thank you xx

Printable View

- Oct 24th 2006, 03:11 PMconcentrate...discriminant?!!!
use the discriminant to find the range of values that K can take if the equation K$\displaystyle x^2$-4x+5-K=0 has no real roots

:eek: someone please help!!! thank you xx - Oct 24th 2006, 03:23 PMPlato
You must solve the following for K.

$\displaystyle b^2 - 4ac < 0 \Rightarrow 16 - 4(k)(5 - k) < 0$ - Oct 24th 2006, 03:29 PMconcentrate...
soo... are you not gonna help me work it out then? or did u jus do so? lol :confused:

- Oct 24th 2006, 03:52 PMPlato
- Oct 24th 2006, 06:26 PMRanger SVO
- Oct 24th 2006, 06:33 PMAfterShock
If someone has made a genuine attempt at a problem, I don't see how helping them obtain an answer is detrimental. That way they are able to complete the other homework problems, which are more than likely similar to the one asked. If they have multiple questions, show how to complete one and let them solve the others. Then, they can show what they learned and can post their solutions to see if they are correct. Or, at the least, providing an obvious hint that that leads to the solution.

- Oct 24th 2006, 07:53 PMJameson
I once tried to debate this point on this site and got threatened :eek: Some people like to learn through getting hints, some like to have the whole solution and work through it. I think the first is more beneficial.

- Oct 24th 2006, 08:05 PMThePerfectHacker
I almost post to the end (the real answer is because I am lazy).

When a person asks a question say on ODE's and I answer it, I do not do it fully because since I know the user needs to be already familar with some concepts and I skip steps and sometimes show the final equation but do not complete it (again because I am lazy).

But I do not think it is helpful when somebody asks how to do a certain differenciation and somebody posts, ..."Use the chain rule". Such a post is a totally useless posts that does not help the reader at all. - Oct 24th 2006, 11:59 PMGlaysher
And in this case I believe this is a test question that the student failed to get right in their recent test (and may have to repeat the same test again)

Just using my psychic powers.

In any case I suspect further explanation is needed

The discriminant is $\displaystyle b^2 - 4ac$

When a quadratic has no real roots then $\displaystyle b^2-4ac<0$

$\displaystyle b = -4$

$\displaystyle a = k$

$\displaystyle c = 5 - k$

Substitute these values into the inequality to get the quadratic inequality

that Plato got. You must solve this (by sketching the graph of $\displaystyle 16 - 4k(5 - k)$ to help) - Oct 25th 2006, 06:14 AMearboth
Hi,

I've got the impression that you and your problem are little bit out of sight.

So let's start again.

1. You've got a quadratic equation in x. I presume that you know that you have to use a special formula to solve this equation:

If the equation is: $\displaystyle ax^2+bx+c=0$, then the solutions are:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{4a}$

2. Compare the given equation with the standard equation. You'll find out:

a = K

b = -4

C = 5-K

Now plug in these terms into the formula and you'll get:

$\displaystyle x=\frac{4\pm\sqrt{16-4*K*(5-K)}}{4K}$

The radicand must be zero or positive (That means greater than zero) to get real values for x. (The radicand is called the dicriminant because it separates the solution into real or complex values)

3. So you have to calculate the values of K so that the dicriminant is zero or positive.

$\displaystyle 16-4*K*(5-K)\geq0$. Expand the LHS of the inequality:

$\displaystyle 4K^2-20K+16\geq0$. Divide by 4

$\displaystyle K^2-5K+4\geq0$. Factorize

$\displaystyle (K-1)(K-4)\geq0$

4. Now you have a product which is gerater or equal zero. A prduct of two factors is positive if both factors have the same sign (More colloquial: +*+ = + or -*-= +). "+" means greater as zero, "-" menas smaller than zero.

So you get:

$\displaystyle K-1\geq 0\ \wedge K-4 \geq 0$ $\displaystyle \vee $ $\displaystyle K-1\le 0\ \wedge \ K-4\le 0$

I'll leave these inequalities for you.

EB - Oct 25th 2006, 12:44 PMconcentrate...