Question

prove that every equation of nth degree has n roots and no more​

Answer 1

Step 1:

Let $f_{n}(x)=a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+...+a_{n-1}x+a_{n}$ ...(i)

i.e., $f_{n}(x) =0$, where $f_{n}(x)=a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+...+a_{n-1}x+a_{n}$ is a polynomial of n degree in x with real coefficient.

Therefore, by the fundamental theorem of algebra, this equation has at least one root.

Step 2:

Let x = α1 be the root of this equation.

Therefore, x - α1 is the factor of $f_{n}(x)$

Let $f_{n}(x) =$ (x - α1) $f_{n-1}(x)$ ...(ii)

where $f_{n-1}(x)$ is a polynomial of degree n - 1 in x with a real coefficient.

Therefore, $f_{n-1}(x) = 0$ has at least one root.

Step 3:

Let x = α2 be the root of this equation $f_{n-1}(x)=0$

Therefore, x - α1 is the factor of $f_{n-1}(x)$

Let $f_{n-1}(x) =$ (x - α2) $f_{n-2}(x)$

where $f_{n-2}(x)$ is a polynomial of degree n - 2 in x with real coefficient.

Step 4:

Substituting the value of $f_{n-1}(x)$ in (ii), we get

$f_{n}(x) =$ (x - α1) (x - α2) $f_{n-2}(x)$

Proceeding in this way, we get

$f_{n}(x) =$ (x - α1) (x - α2)(x - α3).......(x - αn) $f_{n-n}(x)$

$f_{n}(x) =$ (x - α1) (x - α2)(x - α3).......(x - αn) $f_{0}(x)$  ...(iii)

We can write this as

 $a_{0}x^{n}+a_{1}x^{n-1}+...+a_{n-1}x+a_{n}$ = (x - α1) (x - α2)(x - α3).......(x - αn) $f_{0}(x)$

Equating coefficients of $x^{n}$ of both sides

$f_{0}(x) = a_{0}$

Step 5:

Substituting this value in (iii), we get

$f_{n}(x) =$  $a_{0}$ (x - α1) (x - α2)(x - α3).......(x - αn) ...(iv)

Putting x = α1, x = α2, x = α3, .... x = αn in (iv)

$f_{1}$(α1) = 0, $f_{2}$(α2) = 0, $f_{3}$(α3) = 0,... $f_{n}$(αn) = 0

This implies that x = α1, x = α2, x = α3, .... x = αn satisfy the equation $f_{n}(x)=0$

Hence, every polynomial equation of degree n with real coefficients has n roots only.

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