What are the zeros of the function y=2x^2+x-6 ?
cool says 1.y=2x^2+x-6 =2x^2+4x-3x-6 =2x(x+2)-3(x+2) x=3/2;x=-2 2. sq.root of 40/7=2.39 20/3=2.58 _ 3.x=9;x=426 _ sd=underroot sigma (x-x)^2/n =(426-9)^2/13 =115.655
Find the zeros of f(x) = x^2 + 4x - 5 ?
Well first of all you have to expand the brackets out and find 2 numbers which will add to make -5 and times to make 4, which are -1 and 5 as -1+5= 4 and -1x5=-5. so you get (x-1) (x+5). then all you do is the numbers you got is what the zeros are but with different symbols at the front so the answer is 0= +1 or 0= -5. btw both are the answer :) hope that made sense!
How many real zeros does y = 3x4 - 2x2 + x - 3 have?
y = 3x^4 – 2x^2 + x – 3 The Wolfram Alpha site indicates a very ugly solution. Here is just one of the two real zero solutions: x = -(1/6)√{ 4 - 104[2/(9√(74257) - 1231)]^(1/3) + [0.5(9√(74257) -1231)]^(1/3) } – 0.5√( (8/9) + (104/9)[2/(9√(74257) - 1231]^(1/3) - (1/9)[0.5(9√(74257 ) -1231)]^(1/3) + 2/√(4 - 104[2/(9√(74257) - 1231)]^(1/3) + [0.5(9√(74257) - 1231)]^(1/3))) There are four zeroes or roots: 2 real and two complex (imaginary), so there are ONLY TWO REAL zeroes on the real x & y graph. This is a very "ugly" solution that is best done graphically preferably by computer as I did or using a calculator graphing option.
PLEASE HELP. What are the zeros of f(x)=4cos^2x-2 on the interval [0,2pi]?
f(x) = 0 4cos^2 x - 2 = 0 cos^2 x = 2/4 cos x = ± √(1/2) (1/2) = (√2)/2 => cos x = ± √2/2 x = pi/4, 3pi/4, 5pi/4, 7pi/4