Feb 22, 2018 · If x^2+16=0 x=+-sqrt(-16) If ax^2+bx+c=0, then x=1/(2a)(-b+-sqrt(b^2-4ac)) For x^2+16=0 a=1, b=0, c=16 Substituting x=1/(2xx1)(-0+-sqrt(0^2-4xx1xx16)) =1/2xx+-sqrt ...

Feb 17, 2017 · Note that #16=2^4#. #rArr2^(x+2)=(2^4)^x# #rArrcolor(red)(2)^(x+2)=color(red)(2)^(4x)# Since the bases are equal, that is #color(red)(2)# …

c = 16 Substitute these values into the formula above to find that (0, 16) is the vertex point. The expression 16 - x^2 can be factored as a difference of squares. (4 + x)*(4 - x)=0 So, the x …

Jun 1, 2018 · Substitute #x=4sin theta# to turn the denominator into a simple trig function via the identity #sin^2 theta +cos^2 theta=1#. Note that #(dx)/(d theta)=4 cos theta# and that the limits …

For integrals involving the root #sqrt(x^2+a^2),# we can use the substitution #x=atantheta#.Here, #a^2=16, a=4, x=4tantheta#

Mar 23, 2018 · Perform the substitution. #x=4sintheta#, #=>#, #dx=4costhetad theta# #sqrt(16-x^2)=sqrt(16-16sin^2theta)=4costheta#

Mar 11, 2018 · I find that problems like this are solved easier by applying our rules of logarithmic functions. #y = x*sqrt(16-x^2)# We know from the rules of algebra that manipulated one side of …

Sep 10, 2016 · -sqrt(16-x^2)+C Although this is well set up for a shorter, non-trigonometric substitution (see: u=16-x^2), we can make the trigonometric substitution x=4sintheta. Note that …

May 5, 2018 · Here, #I=intx^3sqrt(16-x^2)dx# Let, #x=4sinu=>dx=4cosudu# #and sinu=x/4=>cosu=sqrt(1-sin^2u)=sqrt(1-x^2/16# ...

Apr 9, 2018 · x = -5/4 To factor, we first look to see if there is an immediate factor that is common to all terms. In this case, there is not. Next, we look to find a pair of numbers that when …