Quadratic Equations - How to solve this

[thereddevils]

Homework Statement

Given that the quadratic equation (m^2-3m-4)x^2+(m^2+2)x+12=0 has real roots a and b .Find the set of values of m such that
a<-1<b

Homework Equations

The Attempt at a Solution

By using the fact this equation has two real roots ,b^2-4ac>0.That's all i can see from this problem. Some other hints..

 

[Mark44]

Homework Statement

Given that the quadratic equation (m^2-3m-4)x^2+(m^2+2)x+12=0 has real roots a and b .Find the set of values of m such that
a<-1<b

Homework Equations

The Attempt at a Solution

By using the fact this equation has two real roots ,b^2-4ac>0.That's all i can see from this problem. Some other hints..

That's a good start, so keep going. In your problem, what are a, b, and c?

 

[thereddevils]

That's a good start, so keep going. In your problem, what are a, b, and c?

so i got m^4-44m^2+144m+196>0

i am not sure how does this help to answer the question.

 

[Mark44]

It's also given that a < - 1 and b > -1. Before continuing, make sure that you have copied the problem correctly.

 

[thereddevils]

It's also given that a < - 1 and b > -1. Before continuing, make sure that you have copied the problem correctly.

yes i have correctly copied it down. Which part seems to be missing?

 

[Mark44]

Nothing seemed to be missing, but the fourth degree polynomial doesn't have any integer roots, so the problem seemed more difficult than problems at this level usual are. Maybe you're supposed to find the roots graphically using a graphing calculator.

 

[paulfr]

If you plug into the Quadratic Formula the coefficients [ = f(m)]
you have the solutions as a function of m.
Then apply the two constraints a < -1 < b
And as stated earlier you have b^2 - 4ac giving another equation = f(m)

Also ...
Is it ((m^2)-3m-4) or (m^(2-3m-4)) ?
and
(m^(2+2)) or ((m^2)+2)) ?

 

[Susanne217]

Homework Statement

Given that the quadratic equation (m^2-3m-4)x^2+(m^2+2)x+12=0 has real roots a and b .Find the set of values of m such that
a<-1<b

Homework Equations

The Attempt at a Solution

By using the fact this equation has two real roots ,b^2-4ac>0.That's all i can see from this problem. Some other hints..

As I see it you need to simplify the inequalities

$$m^2-3m-4 < -1$$ and $$m^2+2 > -1$$?? for if this is true I get two real roots!

 

[Unit]

As I see it you need to simplify the inequalities

$$m^2-3m-4 < -1$$ and $$m^2+2 > -1$$?? for if this is true I get two real roots!

The a and b of the quadratic equation are not the same as the roots a and b.

 

[Susanne217]

The a and b of the quadratic equation are not the same as the roots a and b.

Well it should have been ;)

I will look at it again :)

 

[HallsofIvy]

No. Your original problem gave $(m^2-3m-4)x^2+ (m^2+ 2m)x+ 12= 0$ and said that the roots were a and b with a< -1< b. There is no reason to think that $m^2-3m-4< -1< m^2+ m$. Those are the coefficients, not the roots.

(If you are thinking that just because they used the letters "a" and "b", those roots must be the coefficients in "$ax^2+ bx+ c$"- Think again!)

Note instead that if a and b are roots of a quadratic then (x-a) and (x- b) must be factors of the coefficient. Since $(x-a)(x-b)= x^2- (a+b)x+ ab$, the quadratic equation must be of the form $cx^2- c(a+b)x+ abc= 0$ for some number, c. Here, since the leading coefficient, c, is $m^2- 3m- 4$, we must have $(m^2- 3m- 4)(a+ b)= -(m^2+ 2m)$ and $(m^2- 3m- 4)ab= 12$.

 

[ehild]

we must have $(m^2- 3m- 4)(a+ b)= -(m^2+ 2m)$ and $(m^2- 3m- 4)ab= 12$.

There is a typo in the first eq: It is $(m^2- 3m- 4)(a+ b)= -(m^2+ 2)$

Both ab and a+b can be either positive or negative. m^2+2 is always positive. m^2-3m-4= (m-4)(m+1). Check both equations for the possible signs of a+b and ab, if they both can be valid, and for what values of m.

ehild

 

[thereddevils]

There is a typo in the first eq: It is $(m^2- 3m- 4)(a+ b)= -(m^2+ 2)$

Both ab and a+b can be either positive or negative. m^2+2 is always positive. m^2-3m-4= (m-4)(m+1). Check both equations for the possible signs of a+b and ab, if they both can be valid, and for what values of m.

ehild

thanks,

So do i consider the case a<0 , b<0

where (m+1)(m-4)(a+b)<0 since -(m^2+2) is -ve.

so here a+b<0 , (m+1)(m-4)>0 or <0 , but <0 is ignored because in the next case

(m+1)(m-4)ab>0 since 12 is positive.

here (m+1)(m-4) must be >0 since ab>0

m>4, m<-1

When a<0 , b>0

and it look to be for all real numbers of m since i got -1<m<4 here.

 

[ehild]

Sorry, this discussion has not led anywhere. The condition a<-1<b was not used, and this certainly gives some constraints for m. The discriminant has to be also checked if it is really positive. But I could not find the solution yet.

The discriminant is D=(m^2+2)^2-48(m-4)*(m+1). I attach a plot of D in terms of m. It is negative between -7.7 and -1.


ehild

 

[thereddevils]

Sorry, this discussion has not led anywhere. The condition a<-1<b was not used, and this certainly gives some constraints for m. The discriminant has to be also checked if it is really positive. But I could not find the solution yet.

The discriminant is D=(m^2+2)^2-48(m-4)*(m+1). I attach a plot of D in terms of m. It is negative between -7.7 and -1.


ehild

i am quite lost. I am not sure how to continue from where Hallsofivy stopped.

 

[vela]

To get a feel for what the answer might be, you might look at what the curve

$$f(x) = (m^2-3m-4)x^2+(m^2+2)x+12$$

looks like for various values of m. As you vary m, the roots will move around, so you want to figure out where the roots are and how they move as m changes.

For example, the coefficient of x² has a root at m=4. When m=4, f(x) is actually straight line with positive slope that crosses the x-axis at x=-2/3. When m is just a little bit more than 4, like say 4.01, f(x) is still going to be pretty close to the straight line near x=0, which means it'll have a root near x=-2/3, but it will be concave up because (m-4)(m+1)>0 when m>4. That means the other root will be negative, and if m is near 4, it'll be much less than -1. On the other hand, when m is just less than 4, f(x) is concave down, so the other root will be positive.

You should plot the function for m=4 and various values of m, near and away from m=4, to see what I'm talking about. It'll give you a feel for how the roots move as m changes.

Another thing to consider is that f(-1)=0 only when m=2.

 

[thereddevils]

To get a feel for what the answer might be, you might look at what the curve

$$f(x) = (m^2-3m-4)x^2+(m^2+2)x+12$$

looks like for various values of m. As you vary m, the roots will move around, so you want to figure out where the roots are and how they move as m changes.

For example, the coefficient of x² has a root at m=4. When m=4, f(x) is actually straight line with positive slope that crosses the x-axis at x=-2/3. When m is just a little bit more than 4, like say 4.01, f(x) is still going to be pretty close to the straight line near x=0, which means it'll have a root near x=-2/3, but it will be concave up because (m-4)(m+1)>0 when m>4. That means the other root will be negative, and if m is near 4, it'll be much less than -1. On the other hand, when m is just less than 4, f(x) is concave down, so the other root will be positive.

You should plot the function for m=4 and various values of m, near and away from m=4, to see what I'm talking about. It'll give you a feel for how the roots move as m changes.

Another thing to consider is that f(-1)=0 only when m=2.

thanks, but how do i solve it algebraically if this question happens to appear in exams, and graphing calculators are not allower

 

[vela]

I'd be surprised if you could solve this problem algebraically in a straightforward way. In fact, I suspect the point of this problem is to get you to step away from that mentality and to understand how to analyze a function's behavior. This is one of the most useful and powerful skills to develop for problem solving.

Note that what I wrote in the second paragraph of my other post doesn't require a calculator, graphing or not, to figure out. I suggested graphing so you could see how I deduced where the roots would be just based on the concavity of the function.

 

[vela]

I posted this problem in another forum, and one poster came up with a simple and elegant way to solve the problem.

Compare the sign of f(-1) and whether the parabola opens in the upward or downward direction. What combinations will lead to two roots satisfying a<-1<b?

 

[thereddevils]

I posted this problem in another forum, and one poster came up with a simple and elegant way to solve the problem.

Compare the sign of f(-1) and whether the parabola opens in the upward or downward direction. What combinations will lead to two roots satisfying a<-1<b?

Thanks i think i got it,

f(-1)=-3m+6

When the parabola opens upwards, (m^2-3m-4)>0 and f(-1)<0

When the parabola opens downwards, (m^2-3m-4)<0 and f(-1)>0

Either case will generate a<-1<b

so i am left to solve (m^2-3m-4)(-3m+6)<0

Correct >?

 

[science_logic]

correct!

 

[vela]

Thanks i think i got it,

f(-1)=-3m+6

When the parabola opens upwards, (m^2-3m-4)>0 and f(-1)<0

When the parabola opens downwards, (m^2-3m-4)<0 and f(-1)>0

Either case will generate a<-1<b

so i am left to solve (m^2-3m-4)(-3m+6)<0

Correct >?

Yup. Seems really simple in hindsight, huh? ;)