How do you simplify 2 sqrt(3) times 4 sqrt(2)? | Socratic 8sqrt(6) You treat the square root and the natural number differently First you multiply the 2 and 4, and then you do the math inside the square root which would be 3 and 2 Then you put them together, and you get the answer 2sqrt(3) * 4 sqrt(2) = 2 * 4 * sqrt(3 * 2) = 8 sqrt(6)
If z in CC then what is sqrt(z^2)? | Socratic Unless I'm missing something: sqrt (z^2) = z " (primary root) " or +-z " (primary and secpndary roots)" I'm not sure why specifying z in CC is significant
int_ (-a)^adx ( (a^2+x^2)sqrt ( (2a^2+x^2)))=? | Socratic int_ (-a)^ (a) \ 1 ( (a^2+x^2)sqrt (2a^2+x^2)) \ dx = pi (3a^2) We seek the definite integral: I = int_ (-a)^ (a) \ 1 ( (a^2+x^2)sqrt (2a^2+x^2)) \ dx For simplicity, consider the indefinite integral: J = int \ 1 ( (a^2+x^2)sqrt (2a^2+x^2)) \ dx Let us perform a substitution, where we define: x = sqrt (2)a tanu => dx (du) = sqrt (2)a sec^2
How do you solve by completing the square: - 2 = 0? | Socratic x^2+4x+2=0 Remove the 2 from the left side expression by subtracting 2 from both sides: x^2+4x=-2 If the left side is to be a square its non-x term must be the coefficient of x divided by 2 then squared (i e (4 2)^2 = 4); add this amount to both sides: x^2+4x+4=+2 Rewrite the left-hand side as a square: (x+2)^2 = 2 Take the square root of both sides x+2 = +-sqrt(2) Isolate x on the rights