Question

Find the square root of

a^4 - 2a^3 + 3a^2/2 - a / 2 + 1/16

Answer 1

hey i didnt get 3a^2/2 - a / 2 these two!

Answer 2

\[a ^{4} - 2a ^{3} + 3a ^{2}/2 - a/2 + 1/16\]

the / = fraction

write it on paper, i do not know how to insert an fraction

Answer 3

You have to try synthetic division, trial and error. It may turn out this is a perfect square.

Answer 4

\[f(a)=a^4-2a^3+\frac{3a^2}{2}-\frac{a}{2}+\frac{1}{16}=\]
\[=a^4-4a^3\left(\frac{1}{2}\right)+6a^2\left(\frac{1}{2}\right)^2-4a\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4=\]
\[=\sum_{k=0}^4 comb(4,k)\cdot a^{4-k}\left(-\frac{1}{2}\right)^k=
\left(a-\frac{1}{2}\right)^4\]
\[\Rightarrow\quad\sqrt{f(a)}=\left(a-\frac{1}{2}\right)^2=\left(\frac{1}{2}-a\right)^2\]