Is this the discriminate?

**Mukilab** · September 12th 2012, 12:50 PM

My equation is $x^2+2kx-5=0$

I have to find the discriminate for this assuming it has 2 real roots

so I do $b^2-4ac$ is greater than 0

$(2k)^2-4(-5)$ is greater than 0
$k^2$ is greater than -5
k= plus or minus square route of -5

I seem to have done the discriminate well but how can a complex number be the solution for the discriminate since I am supposed to have 2 real roots?

I also encountered the same problem trying to do $x^2+2kx-7=0$ (but with this time the square route of -7). Can anyone help out please? Thanks.

**Plato** · September 12th 2012, 12:56 PM

Originally Posted by Mukilab

My equation is $x^2+2kx-5=0$
I have to find the discriminate for this assuming it has 2 real roots
so I do $b^2-4ac$ is greater than 0
$(2k)^2-4(-5)$ is greater than 0
$k^2$ is greater than -5
k= plus or minus square route of -5
I seem to have done the discriminate well but how can a complex number be the solution for the discriminate since I am supposed to have 2 real roots?
I also encountered the same problem trying to do $x^2+2kx-7=0$ (but with this time the square route of -7). Can anyone help out please? Thanks.

The discriminate is $4k^2+20$ which is positive if $k\ne 0~.$
So it has two real root for any $k~.$

**Mukilab** · September 12th 2012, 12:59 PM

Can you please explain how to get those real roots? The method that I learned from my book clearly isn't working.

My method:
I got the discriminate with no problem
but then I equalled it to 0 (to substitute the greater than 0 sign)
at which point I get k^2=-5.

**emakarov** · September 12th 2012, 01:10 PM

The discriminant is positive for all k. Indeed, k^2 >= -5 holds for all k because k^2 is nonnegative. The fact that the equation k^2 = -5 has only complex roots says that it has no real roots. Therefore, (since the function k^2 + 5 is continuous,) its value is either always positive or always negative. In this case, it is obviously always positive.

**Plato** · September 12th 2012, 01:11 PM

Originally Posted by Mukilab

Can you please explain how to get those real roots? The method that I learned from my book clearly isn't working. My method:
I got the discriminate with no problem
but then I equalled it to 0 (to substitute the greater than 0 sign)
at which point I get k^2=-5.

Consider the equation $ax^2+bx+c=0,~a\ne 0~.$
The discriminate is $\Delta=b^2-4ac~.$ It is called that because it discriminates between real and complex roots.

The roots are $\frac{-b\pm\sqrt{\Delta}}{2a}$ . So if $\Delta>0$ there are two real roots; if $\Delta=0$ there is one real root; and if $\Delta<0$ there are two complex roots.

**Mukilab** · September 12th 2012, 01:13 PM

Originally Posted by emakarov

The discriminant is positive for all k. Indeed, k^2 >= -5 holds for all k because k^2 is nonnegative. The fact that the equation k^2 = -5 has only complex roots says that it has no real roots. Therefore, (since the function k^2 + 5 is continuous,) its value is either always positive or always negative. In this case, it is obviously always positive.

Please elaborate, I don't quite understand what you're saying. Are you saying that my answer through solving the discriminant was right? Or are you saying simply my calculation was right but the answer to the discriminant is something completely different?

If it is the latter, please can you explain how to find the right value for k (that is not a complex number)?

**Plato** · September 12th 2012, 01:16 PM

Originally Posted by Mukilab

Please elaborate, I don't quite understand what you're saying. Are you saying that my answer through solving the discriminant was right? Or are you saying simply my calculation was right but the answer to the discriminant is something completely different?
If it is the latter, please can you explain how to find the right value for k (that is not a complex number)?

Reread all the replies.
Yes we are saying that you method is completely off.

**Mukilab** · September 12th 2012, 01:21 PM

Originally Posted by Plato

Consider the equation $ax^2+bx+c=0,~a\ne 0~.$
The discriminate is $\Delta=b^2-4ac~.$ It is called that because it discriminates between real and complex roots.

The roots are $\frac{-b\pm\sqrt{\Delta}}{2a}$ . So if $\Delta>0$ there are two real roots; if $\Delta=0$ there is one real root; and if $\Delta<0$ there are two complex roots.

Thank you for the explanation however the question specifically said to 'prove that, for all real values of k, the roots of the equation are real and different'.

So surely I have to assume the discriminate is >0? But If I do this, as I have shown above I find that k is greater than a complex number. How can that 'prove' that there are real roots, if I only get a complex number as an outcome?

**Mukilab** · September 12th 2012, 01:23 PM

Originally Posted by Plato

Reread all the replies.
Yes we are saying that you method is completely off.

Can you please explain what I'm doing wrong with the method?

I'm finding b^2-4ac to be k^2+5>0

Where do I go from there? How do I 'prove' that the roots of the equation are real and different?

**Plato** · September 12th 2012, 01:27 PM

Originally Posted by Mukilab

Thank you for the explanation however the question specifically said to 'prove that, for all real values of k, the roots of the equation are real and different'.
So surely I have to assume the discriminate is >0? But If I do this, as I have shown above I find that k is greater than a complex number. How can that 'prove' that there are real roots, if I only get a complex number as an outcome?

For every $k$ , $4k^2+20>0$ . So that are two real roots for every $k$ .

If you still do not get it, you need a sit-down with a live tutor.

**HallsofIvy** · September 12th 2012, 01:27 PM

No, you have NOT shown that "that k is greater than a complex number"- for two reasons. First, by hypothesis, k is a real number so it makes no sense to say that k is "greater than an imaginary number" (which is what you really meant). Second, there is no way to define ">" to make the complex numbers an ordered field.

What you have shown is that $k^2> -5$ which, because a real number, squared, is never negative, is true for any value of k.

**Mukilab** · September 12th 2012, 01:30 PM

Originally Posted by Plato

For every $k$ , $4k^2+20>0$ . So that are two real roots for every $k$ .

If you still do not get it, you need a sit-down with a live tutor.

Thank you, I understand now and I'm sorry for taking so long to understand it, it seems very simple now.

I just had the misconception that I had to do what I did with other equations such as (4k-20>0) which is to simplify to find k (i.e. getting 4k>20.... k>5). I never thought it could be so simple because it is literally just a two line answer.

Once again, sorry for my incompetence. I guess I am making it overcomplicated.

**Plato** · September 12th 2012, 01:31 PM

Originally Posted by Mukilab

Can you please explain what I'm doing wrong with the method?
I'm finding b^2-4ac to be k^2+5>0
Where do I go from there? How do I 'prove' that the roots of the equation are real and different?

In $x^2+2kx-5=0$ , $a=1,~b=2k,~\&~c=-5$ .

So $\Delta=b^2-4ac=(2k)^2-4(1)(-5)=4k^2+20~!$

**Mukilab** · September 12th 2012, 01:39 PM

Just as a last note of clarification.

If I wanted to prove an equation had two real roots 'by finding the discriminant' all I would have to write down is:

(It's discriminant)>0
I.E.
b^2-4ac>0 with the values of the equation substituted in.

Is this correct?

**Plato** · September 12th 2012, 01:54 PM

Originally Posted by Mukilab

Just as a last note of clarification.
If I wanted to prove an equation had two real roots 'by finding the discriminant' all I would have to write down is:
(It's discriminant)>0
I.E.
b^2-4ac>0 with the values of the equation substituted in.

If $\Delta>0$ there are two real roots.

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