# Given a=square root of 3 - i and f(x) = x^4-4x^2+16, prove that f(a)=0.

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### 2 Answers

As a is equal to sqrt 3 - i, we can substitute that for x in the expression f(x) = x^4 - 4x^2 + 16.

f(a) = [(sqrt 3 - i)]^4 - 4*[ (sqrt 3 - i)]^2 + 16

=> f(a) = (3 + i^2 - 2i*sqrt 3)^2 - 4*(3 + i^2 - 2i*sqrt 3) + 16

=> f(a) = (2 - 2i*sqrt 3)^2 - 4*(2 - 2i*sqrt 3) + 16

=> f(a) = 4(1 + (i*sqrt 3)^2 - 2*i*sqrt 3) - 8 + 8i*sqrt 3 + 16

=> f(a) = 4 + 4*i^2*3 - 8*i*sqrt 3 - 8 + 8i*sqrt 3 + 16

=> f(a) = 4 - 12 -8 + 16 -8*i*sqrt 3 + 8i*sqrt 3

=> f(a) = 0

**Therefore for f(x) = x^4 - 4x^2 + 16, f(a) = 0 if a = sqrt 3 - i**

If f(a)=0, then a = sqrt3 - i is the root of the polynomial f(x).

We'll substitute x by a and we'll verify if f(a) = 0

f(a) = a^4 - 4a^2 + 16

f(a) = a^2(a^2 - 4) + 16

We'll re-write the difference of squares a^2 - 4 = (a-2)(a+2)

a = sqrt3 - i

We'll square raise both sides:

a^2 = (sqrt3 - i)^2

We'll expand the square:

a^2 = 3 - 2isqrt3 + i^2, where i^2 = -1

a^2 = 2 - 2isqrt3

f(sqrt3 - i) = (2 - 2isqrt3)(2 - 2isqrt3 - 4) + 16

We'll combine like terms inside brackets:

f(sqrt3 - i) = (2 - 2isqrt3)(-2 - 2isqrt3) + 16

f(sqrt3 - i) = -(2 - 2isqrt3)(2 + 2isqrt3) + 16

We'll write the product as a difference of squares:

f(sqrt3 - i) = -(2^2 - 4*3*i^2) + 16

f(sqrt3 - i) = - (4 + 12) + 16

f(sqrt3 - i) = -16 + 16

f(sqrt3 - i) = 0

**So, f(a) = 0, where a = sqrt3 - i, q.e.d.**