The correct options are
B Both roots of the given equation are rational
C c+a<2b D The equation
ax2+2bx+c=0 has both negative real roots
Let, f(x)=(a+b−2c)x2+(b+c−2a)x+(c+a−2b)
Now, f(1)=a+b−2c+b+c−2a+c+a−2b
⟹f(1)=0
It means that 1 is a root of the given quadratic equation.
Given that, a>b>c
⟹a−c>0 and b−c>0
⟹(a−c)+(b−c)>0
⟹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.
∴ Product of roots should be negative
⟹c+a−2ba+b−2c<0
⟹c+a−2b<0 (∵a+b−2c>0)
⟹c+a<2b
∵a,b,c are rational numbers
∴ Product of roots i.e.c+a−2ba+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, ax2+2bx+c=0
Sum of roots = −2b/a
Product of roots = c/a
∵a,b,c are real
∴ Sum and product of roots are also real.
Hence, we can say that the given quadratic equation has real roots.
Now, ∵ Sum of roots is negative and product of roots is positive.
∴ Both roots of the given quadratic equation will be negative.
Hence, option B, C, D are correct.