# Very hard ans unusual quadratic equations. Need help.

#### Cacapipi2

Hello everybody,
I have 2 problems I cannot solve :
1) (3m-2)x^2 +2mx+3m=0
Can we determine values of m for which this equation has 2 real roots x' and x" such as 0<x'<1<x" ?

2) x^2 -2(2m-5)x+29-20m=0 (1)
a) x' and x" being the roots of the equation, calculate, as a function of m the expression :
y=1/(x'-3) + 1/(x"-3)

b)can we determine m so that y=1 ? So that y=-4/17 ?

c) Find a relation between x' and x" indenpendent of m. Find using this relation the values of the "double roots" (that is when x'=x") of the equation (1).

#### JeffM

Hi. Welcome to math forum.

As our name suggests, we provide help, not answers.

So, with respect to question 1, have you solved the original equation for x? What did you get?

#### Cacapipi2

I really need clues. But let me show you what I have tried to do : for the a) :
We Know that y=1/(x'-3) + 1/(x"-3), so :
y=(x'+x"-6)/(x'x"-3(x'+x")+9)
But if x^2+bx+c has two roots x' and x" then we can express it as (x-x')(x-x").
Thus it is easy to see that x'+x"=-b and x'x"=c. Consequently : x'+x"=2(2m-5) and x'x"=29-20m.
So y=(4m+16)/(-32m+68).

#### HallsofIvy

MHF Helper
If x1 and x2 are roots of ax^2+ bx+ c= 0, then we must have a(x- x1)(x- x2)= ax^2- a(x1+ x2)x+ ax1x2= ax^2+ bx+ c so the sum of the roots is x1+ x2= -b/a and the product is x1x2= c/a.

#### Cacapipi2

Of course, I just ask for clues to the solutions I haven't solved the first exercise. I don't know what you mean by solving the original equation for x (I'm french ) but I have no problem with "classical" quadratic equation and discussing the existence and the sign of the roots as a function of m.

#### JeffM

$(3m - 2)x^2 + 2mx + 3m = 0 \implies x = \dfrac{-\ 2m \pm \sqrt{(2m)^2 - 4(3m - 2)3m}}{2(3m - 2)} \implies$

$x = \dfrac{-\ 2m \pm \sqrt{4m^2 - 4(9m^2 - 6m)}}{2(3m - 2)} = \dfrac{-\ 2m \pm 2\sqrt{6m - 8m^2}}{2(3m - 2)} \implies$

$x = \dfrac{-\ m \pm \sqrt{6m - 8m^2}}{3m - 2}.$

To get two distinct roots requires that $6m - 8m^2 > 0.$ Can you proceed from here. I can't do math in French, j'ai oublie tous depuis longtemps.

EDIT: I mean to get two distinct real roots. And then of course we have further restrictions on those roots, which restrictions imply further restrictions on m.

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#### Cacapipi2

First, we know that m =/=2/3 (because then the equation isn't quadratic anymore and there isn't 2 roots).
Alright so to get 2 distinct roots we have : 6m - 8m^2<0 <=> m(6-8m)>0 <=> 2m(3-4m)>0. So having two distinct roots requieres that 0<m<0.75. But we know that x'+x" = -b/a and x'x"=c/a. So :
x'x">0 <=> 3m/(3m+2) >0 <=> m<0 or m>2/3
x'+x" >0 <=> -2m/(3m-2) >0 <=> 0<m<2/3
Thus it is easy to understand that for 0<m<2/3, the roots' sign are different (x'<0<x") and for 2/3<m<3/4 the roots are negative.
So I guess the answer is no ?

#### Cacapipi2

First, we know that m =/=2/3 (because then the equation isn't quadratic anymore and there isn't 2 roots).
Alright so to get 2 distinct roots we have : 6m - 8m^2<0 <=> m(6-8m)>0 <=> 2m(3-4m)>0. So having two distinct roots requieres that 0<m<0.75. But we know that x'+x" = -b/a and x'x"=c/a. So :
x'x">0 <=> 3m/(3m+2) >0 <=> m<0 or m>2/3
x'+x" >0 <=> -2m/(3m-2) >0 <=> 0<m<2/3
Thus it is easy to understand that for 0<m<2/3, the roots' sign are different (x'<0<x") and for 2/3<m<3/4 the roots are negative.
So I guess the answer is no ?

#### JeffM

You need to consider the possibility that there may be infinitely many possible values of m that result in distinct real roots. Of course it is also possible that there are none. That is the ultimate question, but you certainly should not be deciding that ultimate question on the basis of a single value possible value of m.

To start with, you can conclude that IF ANY VALUES OF m result in the equation having at least one real root, then

$6m - 8m^2 \ge 0 \implies 6m \ge 8m^2 \ge 0 \implies \dfrac{3}{4} \ge m \ge 0.$

What can you conclude from the above given that m is to result in the equation having two distinct roots?

Now we know that the roots of a quadratic equation are symetrically located with respect to the x coordinate of the vertex of the corresponding parabola. If both roots are to fall in the interval (0, 1), then clearly the vertex must also. What does that say about m if it exists?