Simplify square root of 4x^4+36x^2+81

$\sqrt{4 x^{4} + 36 x^{2} + 81}$

Step 1

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$\sqrt{4 {(x^{2})}^{2} + 36 x^{2} + 81}$

Step 2

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$\sqrt{4 u^{2} + 36 u + 81}$

Step 3

Factor using the perfect square rule.

Rewrite $4 u^{2}$ as ${(2 u)}^{2}$ .

$\sqrt{{(2 u)}^{2} + 36 u + 81}$

Rewrite $81$ as $9^{2}$ .

$\sqrt{{(2 u)}^{2} + 36 u + 9^{2}}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$36 u = 2 \cdot (2 u) \cdot 9$

Rewrite the polynomial.

$\sqrt{{(2 u)}^{2} + 2 \cdot (2 u) \cdot 9 + 9^{2}}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 u$ and $b = 9$ .

$\sqrt{{(2 u + 9)}^{2}}$

Step 4

Replace all occurrences of $u$ with $x^{2}$ .

$\sqrt{{(2 x^{2} + 9)}^{2}}$

Step 5

Pull terms out from under the radical, assuming positive real numbers.

$2 x^{2} + 9$