Question

If one root of the quadratic equation \(a(b-c)x^2+b(c-a)x+c(a-b)=0\) is 1, then the other root is

• A. \(\frac{b(c-a)}{a(b-c)}\)
• B. \(\frac{a(b-c)}{c(a-b)}\)
• C. \(\frac{a(b-c)}{b(c-a)}\)
• D. \(\frac{c(a-b)}{a(b-c)}\)

Answer 1

The Correct Option is D

We are given the quadratic equation:

\( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) 

One root is given as \( x = 1 \). Let the other root be \( \alpha \).

By Vieta's formulas, the sum of the roots is given by:

\( 1 + \alpha = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} \)

\( 1 + \alpha = -\frac{b(c-a)}{a(b-c)} \)

Solving for \( \alpha \):

\( \alpha = -\frac{b(c-a)}{a(b-c)} - 1 \)

\( \alpha = \frac{-b(c-a) - a(b-c)}{a(b-c)} \)

\( \alpha = \frac{-b(c-a) - a(b-c)}{a(b-c)} \)

\( \alpha = \frac{-bc + ab - ab + ac}{a(b-c)} \)

\( \alpha = \frac{ac - bc}{a(b-c)} \)

\( \alpha = \frac{c(a-b)}{a(b-c)} \)

Thus, the other root is: \( \frac{c(a-b)}{a(b-c)} \).

Answer 2

Given the quadratic equation \(a(b-c)x^2+b(c-a)x+c(a-b)=0\), where one root is 1, we need to find the other root. 
Let's denote the roots of the equation as \(r_1=1\) and \(r_2\). 
Using Vieta's formulas, the sum and product of the roots for a quadratic equation \(Ax^2+Bx+C=0\) are given by:

\[\text{Sum of roots} = -\frac{B}{A}, \quad \text{Product of roots} = \frac{C}{A}\]

For our equation:

\[A = a(b-c), \quad B = b(c-a), \quad C = c(a-b)\]

Since one of the roots \(r_1\) is 1, the sum of the roots is:

\[1 + r_2 = -\frac{b(c-a)}{a(b-c)}\]

Simplifying to find \(r_2\):

\[r_2 = -\frac{b(c-a)}{a(b-c)} - 1\]

Rewriting 1 as a fraction with the same denominator:

\[r_2 = -\frac{b(c-a)+a(b-c)}{a(b-c)}\]

Simplify the numerator:

\[b(c-a) + a(b-c) = bc-ba+ab-ac = c(b-a)\]

Thus,

\[r_2 = \frac{c(a-b)}{a(b-c)}\]

The other root is \(\boxed{\frac{c(a-b)}{a(b-c)}}\).