For which values of parameter A roots of equation satisfy specific condition

Question

For which values of parameter $a$ roots of equation $(3a+2)x^2+(a-1)x+4a+3=0$ satisfy $x_1<-1<x_2<1$ condition?

My try: \begin{cases} {\Delta} \geq 0 \\f(-1)> 0\\ f(1)<0 \\ \ -1<\frac{-(a-1)}{2(3a+2)}< 1 \end{cases}

I solved this system but I didn't get correct solution. Detailed explanation will be awesome!

Answer 1

Your fourth condition is incorrect.

Hint:

You must consider three cases based on the sign of the leading coefficient i.e. $3a+2$, when it is is positive, negative or zero. The zero case is immediately rejected.

For $3a+2>0$, you must have $f(-1)<0,f(1)>0$ and for $3a+2<0$, you must have $f(-1)>0,f(1)<0$.

Can you finish?

Answer 2

For the quadratic equation $ \ (3a+2)x^2 + (a-1)x + (4a+3) \ = \ 0 \ \ , $ what will make this problem somewhat easier to deal with is that there is a rather narrow range of values of $ \ a \ $ that even permit there to be roots in the interval $ \ (-1 \ , \ 1) \ \ . $ Writing the polynomial in "vertex form" yields $$ (3a \ + \ 2)·\left[ \ x \ + \ \frac{(a \ - \ 1)}{2· (3a \ + \ 2)} \ \right]^2 \ + \ (4a \ + \ 3) \ - \ \frac{(a \ - \ 1)^2}{4· (3a \ + \ 2)} $$ $$ = \ \ (3a \ + \ 2)·\left[ \ x \ + \ \frac{(a \ - \ 1)}{2· (3a \ + \ 2)} \ \right]^2 \ + \ \frac{47a^2 \ + \ 70a \ + \ 23}{4· (3a+2)} $$ $$ = \ \ (3a \ + \ 2)·\left[ \ x \ - \ \frac{(1 \ - \ a)}{2· (3a \ + \ 2)} \ \right]^2 \ + \ \frac{(a \ + \ 1)·(47a \ + \ 23)}{4· (3a+2)}= \ 0 \ \ . $$

We can extract from this expression that

• the parabola representing the quadratic polynomial "opens upward" for $ \ a > -\frac23 \ $ and "opens downward" for $ \ a < -\frac23 \ \ ; $

• the axis of symmetry of the parabola is found at $ \ x > 0 \ $ for $ \ -\frac23 < a < 1 \ $ and at $ \ x < 0 \ $ for $ \ a > 1 \ $ and $ \ a < -\frac23 \ \ $ ; and

• the vertex is located at $ \ y > 0 \ $ for $ \ -1 < a < -\frac23 \ \ , \ \ a > -\frac{23}{47} \ \approx \ -0.489 \ \ $ and at $ \ y < 0 \ $ for $ \ a < -1 \ \ \ , \ \ -\frac23 < a < -\frac{23}{47} \ \ . $

Putting this information together tells us that the parabola has two $ \ x-$intercepts (or the quadratic equation, two roots) only with $ \ -1 < a < -\frac23 \ \ \ , \ \ -\frac23 < a < -\frac{23}{47} \ \ . $

The requirement that the roots satisfy the condition $ \ r_1 < -1 < r_2 < 1 \ \ $ eliminates one of these intervals. For the axis of symmetry $ \ x \ = \ h \ $ with the $ \ x-$intercepts at $ \ x \ = \ h \ \pm \ D \ \ , $ we require for the "larger" root $$ -1 \ < \ h \ + \ D \ < \ 1 \ \ \Rightarrow \ \ -1 - h \ < \ D \ < \ 1 - h \ \ \Rightarrow \ \ 1 + 2h \ > \ h - D \ > \ -1 + 2h \ \ . $$ The condition for the "smaller" root then gives us $ \ -1 + 2h < -1 \ \Rightarrow \ h < 0 \ \ . $ This leaves only the interval $ \ -1 < a < -\frac23 \ \ . $

We will want to check that the larger root remains in the interval $ \ -1 < r_2 < 1 \ \ $ for these values of $ \ a \ \ . $ Our "completion of the square" calculation above has already given us the discriminant of the quadratic polynomial, so we find $$ r_2 \ \ = \ \ \frac{(1 \ - \ a) \ - \ \sqrt{-(a \ + \ 1)·(47a \ + \ 23)}}{2· (3a \ + \ 2)} \ \ = \ \ \frac{(1 \ - \ a) \ - \ \sqrt{\frac{144}{47} \ - \ \frac{1}{47}·(47a \ + \ 35)^2}}{2· (3a \ + \ 2)} \ \ . $$ (The "negative-square-root" solution is the larger root, since the denominator is negative here.) A direct analysis is somewhat daunting, but short of graphing this expression, we can say something about its behavior. As has been said, for $ \ a = -1 \ \ , $ the two $ \ x-$intercepts "merge" into the vertex at $ \ x \ = \ \frac{ 1 \ - \ [-1] }{2· (3·[-1] \ + \ 2)} \ = \ -1 \ \ . $ The discriminant has its maximum at $ \ a \ = \ -\frac{35}{47} \ \approx \ -0.745 \ \ , $ for which $$ r_2 \ \ = \ \ \frac{ ( \ 1 \ - \ [-\frac{35}{47}] \ ) \ - \ \sqrt{\frac{144}{47} }}{2· (3·[-\frac{35}{47}] \ + \ 2)} \ \ = \ \ \frac{ 47 \ + \ 35 \ - \ 12\sqrt{ 47} }{2· ( -105 \ + \ 94)} \ \ \approx \ \ 0.012 \ \ . $$ [Alternatively, for $ \ a \ = \ -\frac34 \ \ , $ the polynomial becomes $ \ -\frac14·x^2 - \frac54·x \ \ , $ so $ \ r_2 \ = \ 0 \ \ . \ ] $
Finally, as $ \ a \ $ approaches $ \ -\frac23 \ \ , $ the quadratic equation asymptotically becomes the linear $ \ -\frac53 · x + \frac13 \ = \ 0 \ \ $ and $ \ r_2 \ \rightarrow \ \frac15 \ \ $ as the other root approaches "negative infinity". So we may conclude that the roots meet the specified condition for $ \ -1 < a < -\frac23 \ \ . $