1 and the positive and negative square roots of 2
(x + 1)(x - 15) gives roots of -1 and 15, and a = 14; (x + 5)(x - 3) gives roots of -5 and 3, and a = 2.
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-3/2
you do 3 plus 5 first then you get 8 you have to find out what minus 7 equals 8
8
(x + 1)(x - 15) gives roots of -1 and 15, and a = 14; (x + 5)(x - 3) gives roots of -5 and 3, and a = 2.
I can't integrate a-x /x-3 ?
(x - 4)(3x^3 + 4) The complex roots are the cube root of negative one third times two to the two thirds power and negative two to the two thirds power divided by the cube root of three
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-3/2
0
you do 3 plus 5 first then you get 8 you have to find out what minus 7 equals 8
Dude, stop trying to cheat on your own maths enrichment task loser.
p(x) = 0 => x4 + 2x3 + x2 + 8x - 12 = 0=> (x + 3)*(x - 1)*(x2 + 4) = 0So the imaginary roots of p(x) are the imaginary roots of x2 + 4 = 0that is x = ±2i
8
This quadratic equation has no real roots because its discriminant is less than zero.
if the question is w^4 = 81 {w raised to the power of 4},Then the four roots are w = {3, -3, 3i, -3i}.The plots on the real-imaginary plane would be the points:(3,0)(-3,0)(0,3)(0,-3)