Question

Prove that every algebraic equation of nth degree has n roots and no more.

Answer 1

Step 1:

Let f_n(x) = α₀xⁿ + α₁x^n-1 + α₂x^n-2 + ..... + α_n-1x + α_n ...... (i)

i.e., f_n(x) = 0, where f_n(x) = α₀xⁿ + α₁x^n-1 + α₂x^n-2 + .... + α_n-1x + α_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 - α₂ 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 - α_n) f_n-n(x) 

f_n(x) = (x - α₁) (x - α₂)(x - α₃).......(x - α_n) f₀(x)    .....(iii)

We can write this as

α₀xⁿ + α₁x^n-1 + ..... + α_n-1x + α_n = (x - α₁) (x - α₂)(x - α₃).......(x - α_n) f₀(x)

Equating coefficients of xⁿ of both sides

 f₀(x) = α₀

Step 5:

Substituting this value in (iii), we get

f_n(x) = α₀(x - α₁) (x - α₂)(x - α₃).......(x - α_n)   ......(iv)

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

f₁(α₁) = 0, f₂(α₂) = 0, f₃(α₃) = 0,... f_n(α_n) = 0

This implies that x = α₁, x = α₂, x = α₃, .... x = α_n satisfy the equation f_n(x) = 0

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