If a,b and c € R, then prove that the roots of the eqn 1/(x-a) + 1/(x-b) + 1/(x-c) =0 are always real and cannot have roots if a=b=c.
To prove that the roots of the equation 1/(x-a) + 1/(x-b) + 1/(x-c) = 0 are always real and cannot have roots if a=b=c, we can follow these steps:
Step 1: Simplify the equation.
Start by finding the common denominator of the three fractions on the left side of the equation. The common denominator is (x-a)(x-b)(x-c). Rewriting the equation with the common denominator, we get:
(x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0
Step 2: Expand and simplify.
Expand the products on both sides of the equation and combine like terms:
x^2 - bx - cx + bc + x^2 - ax - cx + ac + x^2 - bx - ax + ab = 0
3x^2 - 2ax - 2bx - 2cx + ab + ac + bc = 0
Step 3: Rearrange the terms.
Rearrange the equation to isolate the quadratic term and the linear term with respect to x:
3x^2 - 2x(a + b + c) + (ab + ac + bc) = 0
Step 4: Analyze the discriminant.
The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by b^2 - 4ac. In our equation, the coefficient of x^2 is 3, the coefficient of x is -2(a + b + c), and the constant term is (ab + ac + bc). So our discriminant is:
(-2(a + b + c))^2 - 4(3)(ab + ac + bc)
= 4(a + b + c)^2 - 12(ab + ac + bc)
= 4[a^2 + b^2 + c^2 + 2(ab + ac + bc)] - 12(ab + ac + bc)
= 4(a^2 + b^2 + c^2) - 4(ab + ac + bc)
Notice that the discriminant does not depend on x, only on the values of a, b, and c.
Step 5: Prove the discriminant is non-negative.
To show that the roots are always real, we need to prove that the discriminant is non-negative for any values of a, b, and c.
Using the fact that (a - b)^2 + (b - c)^2 + (c - a)^2 ≥ 0 for any real numbers a, b, and c, we can rewrite the discriminant as:
4(a^2 + b^2 + c^2) - 4(ab + ac + bc)
= (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ac + a^2)
= (a - b)^2 + (b - c)^2 + (c - a)^2
Since the sum of squares is always non-negative, the discriminant is non-negative for any values of a, b, and c. Therefore, the roots of the equation are always real.
Step 6: Prove the equation has no roots if a=b=c.
If a=b=c, then the denominator (x - a)(x - b)(x - c) would be equal to zero, which means the equation is undefined and has no roots.
Therefore, we have proved that the roots of the equation 1/(x-a) + 1/(x-b) + 1/(x-c) = 0 are always real and cannot have roots if a=b=c.