Question

$\Huge{\boxed{\sf{Questions}}}$


If x₁ , x₂ , x₃ ....., xₙ are the roots of

xⁿ + ax + b = 0 then the value of

(x₁ - x₂)( x₁ - x₃)(x₁ - x₄)........(x₁ - xₙ)

is equal to ?

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Answer 1

Answer:

$\displaystyle{\implies n.x^{n-1}+a }$

Step-by-step explanation:

Given :

p ( x ) = xⁿ + ax + b = 0

Roots of p ( x )  are x₁ , x₂ , x₃ ....., xₙ .

Now

p ( x₁ ) = (x₁ - x₁ )( x₁ - x₃)(x₁ - x₄)........(x₁ - xₙ)

p ( x₁ ) = 0 ( x₁ - x₃)(x₁ - x₄)........(x₁ - xₙ)

p ( x₁ ) = 0

Now for

(x₁ - x₁ )( x₁ - x₃)(x₁ - x₄)........(x₁ - xₙ)  

Taking limit right side x tends to x₁

$\displaystyle{\implies \lim_{n \to x_1} \frac{x^n+ax+b}{x-x_1} }$

Applying L hospital rule here

$\displaystyle{\implies \lim_{n \to x_1} \frac{n.x^{n-1}+ax^{1-1}+0}{1} }\\\\\\\displaystyle{\implies \lim_{n \to x_1} \frac{n.x^{n-1}+a}{1} }\\\\\\\displaystyle{\implies n.x^{n-1}+a }$

Thus we get answer.

Answer 2

$\Huge{\boxed{\sf{Answer}}}$

The given question:-

p ( x ) = xⁿ + ax + b = 0

x₁ , x₂ , x₃ then till... xₙ

So,

• p ( x₁ ) = (x₁ - x₁ )( x₁ - x₃)(x₁ - x₄)---[(x₁ - xₙ)]

• p ( x₁ ) = 0 ( x₁ - x₃)(x₁ - x₄)--[(x₁ - xₙ)]

• •°• [p ( x₁ ) = 0]

So, then

• (x₁ - x₁ )( x₁ - x₃)(x₁ - x₄)--(x₁ - xₙ)

n x¹ => x^n + ax + b/x - x¹

Let us apply the rule:-

(L hospital rule)

$\begin{lgathered}{\lim_{n \to x_1} \frac{n.x^{n-1}+ax^{1-1}+0}{1} }\\{ \lim_{n \to x_1} \frac{n.x^{n-1}+a}{1} }\\{ n.x^{n-1}+a }\end{lgathered}$