Help with finding a positive and negative root

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July 14th 2010, 06:27 AM
silvercoo

Help with finding a positive and negative root

Hi all..

I've a question which requires to me to find a positive and negative real root for values of p...the equation is...

x^2 +px +8=p

Find the values of p has one positive and one negative root.

Many thanks!
July 14th 2010, 06:47 AM
earboth

Quote:

Originally Posted by silvercoo

Hi all..

I've a question which requires to me to find a positive and negative real root for values of p...the equation is...

x^2 +px +8=p

Find the values of p has one positive and one negative root.

Many thanks!

1. You have to solve the equation for x using the quadratic formula:

$x^2+px+(8-p)=0~\implies~x=-\frac p2 \pm \frac12 \sqrt{p^2+4p-32}$

2. Since $p^2+4p-32 = (p-4)(p+8)$ you know that $p \geq 4$

3. Since the equation should have positive solutions too you know that

$-\frac p2 + \frac12 \sqrt{p^2+4p-32} > 0~\implies~ \frac12 \sqrt{p^2+4p-32} > \frac p2$

4. Square both sides and you'll get $p > 8$
July 14th 2010, 06:50 AM
silvercoo

Hi!

Many thanks for your help..

just wondering, why cant I use b^2 - 4ac >0 ?
July 14th 2010, 07:05 AM
earboth

Quote:

Originally Posted by silvercoo

Hi!

Many thanks for your help..

just wondering, why cant I use b^2 - 4ac >0 ?

That's necessary to get real solutions. But:

Since $p \geq 4$ you know that $-\frac p2$ is a negative number. To get a positive solution the value of the square-root must be greater than that. I used exactly this property to get the inequality.
July 14th 2010, 07:07 AM
silvercoo

Correct, then i would need to equate it to >0

so i should obtain (p+8)(p-4)>0

which would lead to
p<-8 or p>4...

Edit
Pardon me, it should be p>= 4

Edit, again
Nope...its still <-8 or p>4...

sorry about this..
July 14th 2010, 07:24 AM
earboth

Quote:

Originally Posted by silvercoo

Correct, then i would need to equate it to >0

so i should obtain (p+8)(p-4)>0

which would lead to
p<-8 or p>4...

Edit
Pardon me, it should be p>= 4

Edit, again
Nope...its still <-8 or p>4...

sorry about this..

You are correct: There are also positive solutions if $p \leq -8$ .

Sorry for the confusion.
July 14th 2010, 07:29 AM
earboth

Quote:

Originally Posted by silvercoo

...
Edit, again
Nope...its still <-8 or p>4...

sorry about this..

Try p = 5
July 14th 2010, 07:47 AM
silvercoo

Hey thanks. I think i managed to figure it out already..

thanks once again! =)
July 14th 2010, 07:53 AM
running-gag

The product of the roots of the quadratic equation ax²+bx+c=0 is c/a

Therefore to get a positive and a negative root for $x^2+px+(8-p)=0$ you must have a negative product : 8-p < 0 or p > 8