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

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 rootsLet, $$f(x) = (a+b-2c)x^{2}+(b+c-2a)x+(c+a-2b)$$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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