# If a,b,c are positive rational numbers such that a>b>c then tell which of the following statement are correct following quadratic equation

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

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 $$D=b^2-4ac>0$$?

• arent you missing something, option 4 has a different quadratic equation than the one you mentioned in the end – ADITYA PRAKASH Mar 27 at 21:08
• I'll solve it for u – ADITYA PRAKASH Mar 27 at 21:13

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.