Solve Inequalities: Find p Values for p(x^2+x) < 2x^2 + 6x +1

[garytse86]

Can someone help me please with inequalities, I have been attempting this question quite a few times but I still can't get the same answer as the textbook.

What is the set values of p for which p(x^2+x) < 2x^2 + 6x +1 for all real values of x?

Here is my best attempt:

p(x^2+x) < 2x^2 + 6x +1

px^2 + 2p - 2x^2 - 6x -1 < 0

(p-2)x^2 - 6x + 2p -1 < 0

For real roots b^2 - 4ac >= 0

so 6^2 - 4 * (p-2) * (2p-1) >= 0

36 - (8p^2 - 20p + 8) >= 0

-8p^2 + 20p + 28 >= 0


root 1: (-20 + (400 - (4 * -8 * 28)) ^ 1/2) / -16 >= 0
root 1: -20/-16 + (1296^1/2)/16
root 1: 400/256 + 1296/256
= 1696/256

root 2: (-20 - (400 - (4 * -8 * 28)) ^ 1/2) / -16 >= 0
root 2: -20/-16 - (1296^1/2)/16
root 2: 400/256 - 1296/256
= -896/256

That's as far as I have got...

the answer in the textbook is p < -1

Can you please help... Thanks a lot.

Gary.

 

[bogdan]

There's a mistake...

Not "px^2 + 2p - 2x^2 - 6x -1 < 0"...
but...px^2 + p*x - 2x^2 - 6x -1 < 0...
(p-2)*x^2+(p-6)*x-1<0
delta=p^2-12p+36+4p-8...
delta=p^2-8p+28...
I) If p-2>0 then there's no solution...it has a "minimus?"...
II) If p-2<=0 then...
delta<0
p^2-8p+28<0
p^2-8p+16+12<0
(p-4)^2+12<0...which is impossible...so there's no such p...
Are you sure the text is correct ?

 

[garytse86]

i am really sorry, the question should be:

p(x^2+2) < 2x^2 + 6x +1

not

p(x^2+x) < 2x^2 + 6x +1

 

[bogdan]

p(x^2+2) < 2x^2 + 6x +1

Then we have...

(p-2)*x^2-6*x+(2p-1)<0
Evidently p-2<0...the function has a "maximum"...
delta=36-4(2p-1)(p-2)...
delta=36-4(2p^2-5p+2)...
delta=36-8p^2+20p-8...

delta<0...no real solutions...
-8*p^2+20*p+28<0...
2*p^2-5*p-7>0...
dp->delta for this ecuation...
dp=25+56...
dp=81...
p1=(5-9)/4=-1...
p2=(5+9)/4=7/2...
because a in this ecuation is >0...p is in (-inf,-1)U(7/2,+inf)...
now...because p-2<0...p<2...results that p<-1...
Got it ?

 

[StephenPrivitera]

Evidently p-2<0...the function has a "maximum"...

How do you know this?

 

[bogdan]

let f(x)=a*x^2+b*x+c...
f'(x)=2*a*x+b
f''(x)=2*a
where f'(x)=0 you'll have an "extremum"...
f'(x0)=2*a*x0+b=0...x0=-b/(2*a)...
f''(x0)=2*a...if f''(x0)>0 then in x0 the concavity is upward...
else it is downward...so...if a>0 the function looks like this \/...and if a<0 then it looks like this /\...got it now ?

 

[garytse86]

bogdan, sorry this is the first time I have done this, when you said:

delta<0...no real solutions...
-8*p^2+20*p+28<0...

Then shouldn't the last line be:

-8*p^2+20*p+28>0...

In the textbook it showed the calculations exactly the same way you did, that's what I don't get.

Would you mind explaining this?

 

[HallsofIvy]

Here's how I would do this problem:

p(x^2+2) < 2x^2 + 6x +1 is the same as

px^2+ 2p< 2x^2+ 6x+ 1 or (2-p)x^2+ 6x+ (1-2p)> 0 for all x.

If this is never 0, then it's discriminant must be negative:

36- 4(2-p)(1-2p)< 0

(The inequality is <0 so that the discrimant is negative and there are no real roots.)

-8p^2+ 20p+ 28< 0
2p^2- 5p- 7> 0

2p^2- 5p- 7= 0 has roots -1 and 7/2.

2p^2- 5p- 7> 0 for x< -1 and x> 7/2.

Putting x= -2 (less than -1) into the original inequality, we find that p(x^2+2) < 2x^2 + 6x +1 for p< -1- the result we want.
Putting x= 3 (greater than 7/2) into the original inequality, we find that p(x^2+2) > 2x^2 + 6x +1 for p> 7/2- not what we want.

The original inequality is true for all x as long as p< -1.

 

[bogdan]

delta=36-8p^2+20p-8...

delta<0...no real solutions...
-8*p^2+20*p+28<0...
if delta<0 then the function doesn't intersect OX...and because it looks like /\...it must be under OX for every x...got it ?

 

[garytse86]

Question: What is the set values of p for which p(x^2+x) < 2x^2 + 6x +1 for all real values of x?

Quote: for all real values of x?

Then b^2 - 4ac > 0 should be true not

b^2 - 4ac < 0

thats what I don't understand, and I really appreciate your help!

 

[HallsofIvy]

Question: What is the set values of p for which p(x^2+x) < 2x^2 + 6x +1 for all real values of x?

Quote: for all real values of x?

Then b^2 - 4ac > 0 should be true not

b^2 - 4ac < 0

thats what I don't understand, and I really appreciate your help!

Well, first, you are back to "x^2+ x" which you told us was incorrect. I'm going to assume you meant p(x^2+2) < 2x^2 + 6x +1.

As far as "b^2- 4ac" is concerned you have it backwards.

If ax^2+ bx+ c> 0 (or <0) for all x, then ax^2+ bx+ c must have NO real solutions. That means the two roots that any quadratic has must both be complex: the discriminant, b^2- 4ac must be negative.

 

[bogdan]

if that delta is non-negative then the ecuation would have at least one solution...that means the inequality would be <=...not <...
...draw the graphic of the function...and you'll see...try it with both a<0...and a>0...and delta<0...and delta>0...and it will become clear...

 

[garytse86]

HallsofIvy, ax^2+bx+c > 0 does not mean both roots are not real.

consider x^2 + 2x + 3 > 0, the roots are not complex.

 

[HallsofIvy]

I may not be very bright, but I can at least use the quadratic formula: the roots of x^2 + 2x + 3 = 0 are (-2+/- sqrt(4-12))/2=
-1 +/- sqrt(2)i and definitely ARE complex.

Saying that ax^2+ bx+ c> 0 (or ax^2+ bx+ c< 0) for all x means that it is never EQUAL to 0 for any real x. The roots of the equation MUST be complex.

 

[garytse86]

sorry that wasn't meant to be offensive.

but consider x^2 + 3x + 2 = 0, the roots are real.

 

[HallsofIvy]

Yes, the roots of x^2 + 3x + 2 = (x+2)(x+1)= 0 are x= -1 and x= -2. It follows FROM THAT that x^2+ 3x+ 2< 0 for -2< x< -1 and that
x^2+ 3x+ 2> 0 for x< -2 or x> -1.

The point, once again, is that the roots of ax^2+ bx+ c= 0 are (by definition) places where the value is 0. If the ax^2+ bx+ c> 0 or ax^2+ bx+ c< 0 for ALL X, then there is NO place where the value is 0 ans so no real value of x for which the equation is satisfied: the roots MUST be complex and b^2- 4ac must be negative.