If (a +b+c)>0 and a0 b c, then the equationa(x -b)(x -c)+ b(x- c)(x- a

1.	If (a +b+c)>0 and a0 b c, then the equationa(x -b)(x -c)+ b(x- c)(x- a)+c(x - a)(x - b) 0 has(A) real and distinct roots(C) product of roots is negative(B) roots are imaginary(D) product of roots is positive
Answer» The roots are real a(x−b)(x−c)+b(x−c)(x−a)+c(x−a)(x−b)=0a(x−b)(x−c)+b(x−c)(x−a)+c(x−a)(x−b)=0 ⟹a[x2−(b+c)x+bc]+b[x2−(c+a)x+ca]+c[x2−(a+b)x+ab]=0⟹a[x2−(b+c)x+bc]+b[x2−(c+a)x+ca]+c[x2−(a+b)x+ab]=0 ⟹(a+b+c)x2−2(ab+bc+ca)x+3abc=0⟹(a+b+c)x2−2(ab+bc+ca)x+3abc=0 D=4(ab+bc+ca)2−4(a+b+c)(3abc)D=4(ab+bc+ca)2−4(a+b+c)(3abc) =4(a2b2+b2c2+c2a2+2a2bc+2ab2c+2abc2−3a2bc−3ab2c−3abc2)=4(a2b2+b2c2+c2a2+2a2bc+2ab2c+2abc2−3a2bc−3ab2c−3abc2) =4(a2b2+b2c2+c2a2−a2bc−ab2c−abc2)=4(a2b2+b2c2+c2a2−a2bc−ab2c−abc2) =2(2a2b2+2b2c2+2c2a2−2a2bc−2ab2c−2abc2)=2(2a2b2+2b2c2+2c2a2−2a2bc−2ab2c−2abc2) =2[(ab−bc)2+(bc−ca)2+(ca−ab)2]=2[(ab−bc)2+(bc−ca)2+(ca−ab)2] ⟹D≥0∀a,b,c∈R⟹D≥0∀a,b,c∈R Ifx=a=b=cx=a=b=c, we haveD=0D=0

If (a +b+c)>0 and a0 b c, then the equationa(x -b)(x -c)+ b(x- c)(x- a)+c(x - a)(x - b) 0 has(A) real and distinct roots(C) product of roots is negative(B) roots are imaginary(D) product of roots is positive

Answer»

The roots are real

a(x−b)(x−c)+b(x−c)(x−a)+c(x−a)(x−b)=0a(x−b)(x−c)+b(x−c)(x−a)+c(x−a)(x−b)=0

⟹a[x2−(b+c)x+bc]+b[x2−(c+a)x+ca]+c[x2−(a+b)x+ab]=0⟹a[x2−(b+c)x+bc]+b[x2−(c+a)x+ca]+c[x2−(a+b)x+ab]=0

⟹(a+b+c)x2−2(ab+bc+ca)x+3abc=0⟹(a+b+c)x2−2(ab+bc+ca)x+3abc=0

D=4(ab+bc+ca)2−4(a+b+c)(3abc)D=4(ab+bc+ca)2−4(a+b+c)(3abc)

=4(a2b2+b2c2+c2a2+2a2bc+2ab2c+2abc2−3a2bc−3ab2c−3abc2)=4(a2b2+b2c2+c2a2+2a2bc+2ab2c+2abc2−3a2bc−3ab2c−3abc2)

=4(a2b2+b2c2+c2a2−a2bc−ab2c−abc2)=4(a2b2+b2c2+c2a2−a2bc−ab2c−abc2)

=2(2a2b2+2b2c2+2c2a2−2a2bc−2ab2c−2abc2)=2(2a2b2+2b2c2+2c2a2−2a2bc−2ab2c−2abc2)

=2[(ab−bc)2+(bc−ca)2+(ca−ab)2]=2[(ab−bc)2+(bc−ca)2+(ca−ab)2]

⟹D≥0∀a,b,c∈R⟹D≥0∀a,b,c∈R

Ifx=a=b=cx=a=b=c, we haveD=0D=0