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!
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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: 2. Since you know that 3. Since the equation should have positive solutions too you know that 4. Square both sides and you'll get
Hi! Many thanks for your help.. just wondering, why cant I use b^2 - 4ac >0 ?
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 you know that 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.
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..
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 . Sorry for the confusion.
Originally Posted by silvercoo ... Edit, again Nope...its still <-8 or p>4... sorry about this.. Try p = 5
Hey thanks. I think i managed to figure it out already.. thanks once again! =)
The product of the roots of the quadratic equation ax˛+bx+c=0 is c/a Therefore to get a positive and a negative root for you must have a negative product : 8-p < 0 or p > 8
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