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  • How do you simplify the square root of 27 4? - Socratic
    Get the squares out from under the root Explanation: #sqrt(27 4)=sqrt27 sqrt4=sqrt(3^2*3) sqrt(2^2)=(sqrt(3^2)*sqrt3) (sqrt(2^2))=#
  • 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)
  • What is the Square root 20 - square root 5 + square root 45? - Socratic
    We first simplify #sqrt20# and #sqrt45# terms by prime factorisation: #sqrt20 = sqrt(2^2 *5)= color(bluw)(2sqrt5#
  • #sqrt((48x^4))#? - Socratic
    #color(blue)sqrt(48)*sqrt(x^4)# We can factor a perfect square out of #sqrt48# We can factor out a #16# and #3# We would get: #color(blue)sqrt16*color(blue)sqrt3*sqrt(x^4)# (Blue terms are equal to #sqrt48#) #sqrt16# simplifies to #4#, we cannot factor #sqrt3# any further, and #sqrt(x^4)# would simply be #x^2# We have: #4sqrt3*x^2#
  • How do you find the exact square root of 39? - Socratic
    How do you prove that square root 15 is irrational? You can find a succession of rational approximations for #sqrt(39)# using a Newton Raphson type method Typically you would start with an approximation #a_0# and iterate using a formula like: #a_(i+1) = (a_i^2 + n) (2a_i)# where #n = 39# is the number you are trying to approximate the square
  • What is the square root of 32 over square root of 8? - Socratic
    We can rewrite this expression as: #sqrt(32) sqrt(8) => sqrt(8 * 4) sqrt(8)# Now, we can use this rule for radicals to rewrite the numerator and complete the simplification:
  • How do you simplify #root4 (40)#? - Socratic
    2 514866859 By calculator root4(40)=2 514866859
  • How do you find the discriminant of -x^2-x=4 and use it to . . . - Socratic
    The equation has no real roots and two imaginary roots The "discriminant" that the question is referring to is the discriminant of the quadratic formula, which is b^2-4ac I've highlighted it in the quadratic formula here in red: x=(-b+-sqrt(color(red)(b^2-4ac))) (2a) We know that by the fundamental theorem of algebra that any quadratic has exactly 2 roots which can be real or imaginary If
  • What is the square root of (-12)^2? - Socratic
    In this case, since the principal square root or a positive number is the non-negative square root, the answer is #12# Note that for non-negative real #n#, the symbol #sqrtn# always refers to the principal square root The definition of a square root is: #a# is a square root of #b# if and only if #a^2 = b# So every positive number has 2





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