Quadratic EquationHard
Question
Both the roots of the equation (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b) = 0 are always
Options
A.positive
B.negative
C.real
D.None of these
Solution
(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0
⇒ 3x2 - 2(a + b + c)x + (ab + bc + ca) = 0
Now, discriminant
= 4(a + b + c)2 - 12(ab + bc + ca)
= 4{a2 + b2 + c2 - ab - bc - ca}
= 2{(a - b)2 + (b - c)2 + (c - a)2}
which is always positive.
Hence, both roots are real.
⇒ 3x2 - 2(a + b + c)x + (ab + bc + ca) = 0
Now, discriminant
= 4(a + b + c)2 - 12(ab + bc + ca)
= 4{a2 + b2 + c2 - ab - bc - ca}
= 2{(a - b)2 + (b - c)2 + (c - a)2}
which is always positive.
Hence, both roots are real.
Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
If the value of quadratic trinomial $ax^{2} - bx + c$ is an integer for $x = 0,x = 1$ and $x = 2$, then the valueof the ...If x − 2 is a common factor of x2 + ax + b and x2 + cx + d, then -...Let g(x) = cos x2, f(x) = x and α, β (α<β) be the roots of the quadratic equation 18x2...If α ≠ β but α2 = 5α - 3 and β3 = 5β - 3 then the equation having α/β and ...The roots of the equation log2 (x2 − 4x + 5) = (x − 2) are -...