Help with finding a positive and negative root

• Jul 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!
• Jul 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:

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

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

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

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

4. Square both sides and you'll get $\displaystyle p > 8$
• Jul 14th 2010, 06:50 AM
silvercoo
Hi!

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

Originally Posted by silvercoo
Hi!

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

That's necessary to get real solutions. But:

Since $\displaystyle p \geq 4$ you know that $\displaystyle -\frac p2$ 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.
• Jul 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

p<-8 or p>4...

Edit
Pardon me, it should be p>= 4

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

• Jul 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

p<-8 or p>4...

Edit
Pardon me, it should be p>= 4

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

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

Sorry for the confusion.
• Jul 14th 2010, 07:29 AM
earboth
Quote:

Originally Posted by silvercoo
...
Edit, again
Nope...its still <-8 or p>4...

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