How do you simplify #sqrt(–(–7)^2)#? Algebra Radicals and Geometry Connections Simplification of Radical Expressions 2 Answers NJ Apr 17, 2018 #=>7i# Explanation: #sqrt(-(-7)^2)# #=sqrt(-49)# #=sqrt((49)*(-1))# #=sqrt(49)*sqrt(-1)# #=sqrt(7^2)*sqrt(-1)# #=(7)*(i)# #=7i# Answer link Monzur R. Apr 17, 2018 #sqrt(-(-7)^2)=+-7i# Explanation: #sqrt(-(-7)^2)=sqrt(-49)=sqrt(-1)sqrt(49)=+-i7=+-7i# Over #RR#, we would only take the positive square root of #sqrt(a^2)#, but the square root function is multi-variable over #CC#, so we take both roots. Answer link Related questions How do you simplify radical expressions? How do you simplify radical expressions with fractions? How do you simplify radical expressions with variables? What are radical expressions? How do you simplify #root{3}{-125}#? How do you write # ""^4sqrt(zw)# as a rational exponent? How do you simplify # ""^5sqrt(96)# How do you write # ""^9sqrt(y^3)# as a rational exponent? How do you simplify #sqrt(75a^12b^3c^5)#? How do you simplify #sqrt(50)-sqrt(2)#? See all questions in Simplification of Radical Expressions Impact of this question 1459 views around the world You can reuse this answer Creative Commons License