Multiple choice

If the roots of equation $ a\left( b-c \right) { x }^{ 2 }+b\left( c-a \right) x+c\left( a-b \right) =0$ be equal, then

  1. $2b= a + c$
  2. $b^2 = ac$
  3. $\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$

  4. None of these.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For equal roots, the discriminant b^2 - 4ac = 0. Here, A = a(b-c), B = b(c-a), C = c(a-b). B^2 - 4AC = 0 leads to [b(c-a)]^2 - 4[a(b-c)][c(a-b)] = 0. This simplifies to b^2(c-a)^2 + 4ac(b-c)(a-b) = 0. Expanding and rearranging leads to the condition 2/b = 1/a + 1/c.

AI explanation

Let one root be 1, which makes x = 1 a solution to the given equation a(b-c)x^2 + b(c-a)x + c(a-b) = 0. Substituting x = 1 yields a(b-c) + b(c-a) + c(a-b) = 0, which simplifies to ab - ac + bc - ab + ac - bc = 0 and confirms 1 is always a root. Since the roots are equal, both roots must be exactly 1. The product of the roots for this equation is c(a-b) / a(b-c), so c(a-b) / a(b-c) = 1. Expanding this gives ac - bc = ab - ac, and rearranging results in 2ac = b(a+c). Dividing the entire equation by abc yields the relationship 2/b = 1/c + 1/a.

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