Multiple choice

If ${ x }{ 1 },{ x }{ 2 },{ x }{ 3 },..,{ x }{ n }$ are the roots of the equation $x^n+ax+b=0,$ then the value of $\left( { x }{ 1 }-{ x }{ 2 } \right) \left( { x }{ 1 }-{ x }{ 3 } \right) \left( { x }{ 1 }-{ x }{ 4 } \right) ...\left( { x }{ 1 }-{ x }{ n } \right) $ is equal to

  1. $n{ { x }_{ 1 } }^{ n-1 }+a$
  2. $n{ { x }_{ 1 } }^{ n-1 }$
  3. $nx-1+b$
  4. $n{ { x }_{ 1 } }^{ n-1 }+b$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If x1, x2... are roots of P(x) = x^n + ax + b = 0, then P(x) = (x-x1)(x-x2)...(x-xn). The derivative P'(x) = nx^(n-1) + a. Also P'(x) = sum over i of [P(x)/(x-xi)]. At x=x1, P'(x1) = (x1-x2)(x1-x3)...(x1-xn). Thus the product is nx1^(n-1) + a.

AI explanation

Let P(x) = x^n + ax + b. Since x_1 is a root of the polynomial, we can write x_1^n + ax_1 + b = 0. The derivative of the polynomial is P'(x) = nx^(n-1) + a. For a polynomial with distinct roots, the product (x_1 - x_2)(x_1 - x_3)...(x_1 - x_n) is exactly equal to P'(x_1). Therefore, substituting x_1 into the derivative gives the result nx_1^(n-1) + a.

Keep practicing — related questions