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I am solving following question based on quadratic equation

If $a,b,c$ are positive rational numbers such that $a>b>c$ and the quadratic equation $(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0$ has a root in the interval $(-1,0)$ then which of the following statements are true ?

  1. $b+c>a$
  2. $c+a<2b$
  3. both roots of the given equation are rational
  4. the equation $ax^2+2bx+c=0$ has both negative real roots.

My Approach

First I calculated discriminant of the given quadratic equation which turns out to be $3(b-c)$ (This proves statement 3).

So root 1 $r_1$ is

$$r_1 = \frac{-b-c+2a+3b-3c}{2(a+b-2c)}=1$$

So root 2 will be

$$\frac{c+a-2b}{a+b-2c}$$

As it is mentioned that one root will in $(-1,0) $ so $\frac{c+a-2b}{a+b-2c}$ will be that root. So

$$ -1<\frac{c+a-2b}{a+b-2c}<0 \\ -a-b+2c<c+a-2b<0 $$

Solving first half of the above inequality i.e. $c+a-2b<0$ will prove statement 2 to be true.

Solving another half of the inequality i.e. $-a-b+2c < c+a-2b$ will prove statement 1 to false as our results are $b+c<2a$.

But I am not able to find the reasoning for fourth statement. My work for proving 4th statement to true:

As $a,b,c$ are all positive and sum of the root for $ax^2+2bx+c$ is $\alpha+\beta=-2b/a$. This proves that at least one of the root is negative. The product of the root is $\alpha\beta=c/a$ as $c/a$ is positive this states that both the roots are negative.

if the discriminant of the $ax^2+2bx+c$ is > 0 only then this equation will have real roots. How do I prove that the discriminant $D=b^2-4ac>0$?

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  • $\begingroup$ arent you missing something, option 4 has a different quadratic equation than the one you mentioned in the end $\endgroup$ – ADITYA PRAKASH Mar 27 at 21:08
  • $\begingroup$ I'll solve it for u $\endgroup$ – ADITYA PRAKASH Mar 27 at 21:13
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Building up on your work :- You have found that $$2b>a+c... (1)$$$$2a>b+c...(2)$$Note that $ax^2+2bx+c$ has all positive coefficients , there fore the parabola of this quadratic equation will have its vertex in the negative X - direction (${-b\over 2a} <0$ and we will have an upturned parabola ($a>0$) and Y-intercept of the parabola will be positive ($c>0$).

Analysing the above characteristics, there are only two contending parabolas ($A$ and $B$ in the diagram)

Note that (1) can be rearranged to $a-2b+c <0$ which is nothing but f(-1) or the y-value at x=-1 , therefore case B is correct and the equation has both negative roots. enter image description here

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