Simplify $\sqrt{a^6 + 2a^4b^2 + a^2b^4}$

Question

Simplify $\sqrt{a^6 + 2a^4b^2 + a^2b^4}$ for $a < 0$

I've almost got it but I have a question about the answer. This is my solution:

$$\sqrt{a^6 + 2a^4b^2 + a^2b^4}$$ $$\sqrt{a^2 (a^4 + 2a^2b^2 + b^4)}$$ $$a \cdot \sqrt{a^4 + 2a^2b^2 + b^4}$$ $$a^2 = t$$ $$a \cdot \sqrt{t^2 + 2b^2t + b^4}$$ $$a \cdot \sqrt{(t+b^2)^2}$$ $$a \cdot |t+b^2|$$ $$a \cdot |a^2+b^2|$$

Now splitting the cases for when $a^2 + b^2 < 0$ vs $a^2 + b^2 > 0$ gives $$a \cdot (a^2+b^2)$$ $$-a \cdot (a^2 + b^2)$$

Now, my textbook only gives the solution $$-a\cdot(a^2+b^2)$$ Why?

Answer 1

When you moved $a^2$ out from square root, you forgot to take absolute value:

$$\sqrt{a^2 \cdot (a^4+2a^2b^2+b^4)}=|a| \sqrt{a^4+2a^2b^2+b^4}$$

and since $a < 0$, we have $$|a| = -a$$

Answer 2

since $a<0$ $$\sqrt { a^{ 6 }+2a^{ 4 }b^{ 2 }+a^{ 2 }b^{ 4 } } =\sqrt { { a }^{ 2 }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right) }^{ 2 } } =\left| a\left( { a }^{ 2 }+{ b }^{ 2 } \right) \right| =\left| a \right| \left| \left( { a }^{ 2 }+{ b }^{ 2 } \right) \right| =-a\left( { a }^{ 2 }+{ b }^{ 2 } \right) \\ $$

Answer 3

In general, $$\sqrt{t^2}=|t|=t\quad\text{if $\ t\ge 0$}$$ $$\sqrt{t^2}=|t|=-t\quad\text{if $\ t\lt 0$}$$

In your case, $$\sqrt{a^6+2a^4b^2+a^2b^4}=\sqrt{a^2(a^2+b^2)^2}=\sqrt{a^2}\sqrt{(a^2+b^2)^2}$$ which is equal to $$-a(a^2+b^2)$$ since $a\lt 0$ and $a^2+b^2\gt 0$.