The equation x^{4}+4x^{2}+3=0 is a bi-quadratic equation. In this post, we will first factorise x^{4}+4x^{2}+3 and then solve the equation x^{4}+4x^{2}+3=0.

## How to Factorise x^{4}+4x^{2}+3

**Question: Factorize x ^{4}+4x^{2}+3**.

**Solution:**

To factorize x4+4×2+3, our aim is to express it in the form of a^{2}-b^{2}. For that, we need to add and subtract 1 to both sides. So we have

x^{4}+4x^{2}+3

= x^{4}+4x^{2}+3 + 1 -1

= x^{4}+4x^{2}+4 – 1

= (x^{2})^{2}+2**⋅**x^{2}⋅2+2^{2} – 1^{2}

= (x^{2}+2)^{2} – 1^{2} as we know that a^{2}+2ab+b^{2} = (a+b)^{2}

= (x^{2}+2-1) (x^{2}+2+1) using the formula a^{2}-b^{2} = (a-b)(a+b)

Simplifying the above, we obtain that

x^{4}+4x^{2}+3 = (x^{2}+1) (x^{2}+3) …(∗)

Thus, the factorization of x^{4}+4x^{2}+3 is given by x^{4}+4x^{2}+3 = (x^{2}+1) (x^{2}+3).

**You can read:** How to solve linear equations

## Solve x^{4}+4x^{2}+3=0

**Question: Solve the equation x ^{4}+4x^{2}+3=0**.

**Solution:**

x^{4}+4x^{2}+3=0

⇒ (x^{2}+1) (x^{2}+3)=0, follows from above (∗)

We know that if the product of two numbers is zero, then the numbers are individually zero. Thus, we have

x^{2}+1=0 or x^{2}+3=0

⇒ x^{2}-i^{2}=0 or x^{2}-3i^{2}=0 where i=√-1 is an imaginary complex number, so i^{2}=-1.

⇒ x^{2} = i^{2} or x^{2} = 3i^{2}

Taking square root on both sides, we get that

x = i, -i or x = $\sqrt{3}$i, -$\sqrt{3}$i

So the solutions of x^{4}+4x^{2}+3=0 are given by x=±i, ±$\sqrt{3}$i.

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## FAQs

### Q1: What is the factorisation of x^{4}+4x^{2}+3?

Answer: The factorisation of x^{4}+4x^{2}+3 is equal to (x^{2}+1) (x^{2}+3).