Answer:
[tex]\left(4-3\pi \right)^2 \text{ is a perfect square trinomial}[/tex]
[tex]\left(4-3\pi \right)^2=16-24\pi +9\pi ^2[/tex]
[tex]\sqrt{16-24\pi +9\pi ^2}\approx 5.42[/tex]
[tex]\sqrt{(4 -3\pi )^2} = -4 + 3\pi[/tex]
[tex]\sqrt{(4-3\pi )^2} = |4-3\pi|[/tex]
State that this is wrong, it cannot be negative
[tex]\sqrt{(4 -3\pi )^2} = 4 - 3\pi[/tex]
So,
[tex]\sqrt{(4 -3\pi )^2} \neq 4 - 3\pi[/tex]
State that [tex]\sqrt{a^2}=|a|[/tex]
It is [tex]a[/tex] assuming that a > 0
But we don't know if a is greater than 0.
