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Question

If $$a, b, c$$ are positive rational number 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


A
b+c>A
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B
c+a<2b
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C
Both roots of the given equation are rational
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D
The equation ax2+2bx+c=0 has both negative real roots
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Solution

The correct options are
B Both roots of the given equation are rational
C $$c+a < 2b$$
D The equation $$ax^2+2bx+c=0$$ has both negative real roots
Let, $$f(x) = (a+b-2c)x^{2}+(b+c-2a)x+(c+a-2b)$$

Now, $$f(1)=a+b-2c+b+c-2a+c+a-2b$$
$$\implies f(1)=0$$
It means that 1 is a root of the given quadratic equation.

Given that, $$a>b>c$$
$$\implies a-c>0 \ and\  b-c>0$$
$$\implies (a-c)+(b-c) >0$$
$$\implies a+b-2c >0$$

We got one positive root of the given quadratic equation i.e. 1 and the other root lies between the interval (-1,0). So, the other root is negative.
$$\therefore$$ Product of roots should be negative
$$\implies \frac{c+a-2b}{a+b-2c} <0$$
$$\implies c+a-2b<0$$                           $$(\because a+b-2c>0)$$
$$\implies c+a<2b$$

$$\because a,b,c \ $$ are  rational  numbers
$$\therefore$$ Product of roots $$i.e. \frac{c+a-2b}{a+b-2c}$$ is rational number and we know that one root is 1 i.e. rational. So, the other root must be a rational number.

For, $$ax^{2}+2bx+c=0$$

Sum of roots = $$-2b/a$$
Product of roots = $$c/a$$
$$\because a,b,c \ $$ are real
$$\therefore $$ Sum and product of roots are also real.
Hence, we can say that the given quadratic equation has real roots.

Now, $$\because $$ Sum of roots is negative and product of roots is positive.
$$\therefore$$ Both roots of the given quadratic equation will be negative.

Hence, option B, C, D are correct.

Mathematics

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