find roots of w^(4)+2^(3)+3^(2)+4w-52=0
ok, I read
w^4 + 8 + 9 + 4w - 52 = 0
w^4 + 4w -35=0
No easy way to solve this,
Wolfram shows two real solutions
w = -2.5955 and w = 2.25747
http://www.wolframalpha.com/input/?i=w%5E4+%2B+4w+-35%3D0
notice you also have 2 complex roots.
Assuming the usual careless typing of posters, I read it as
w^4+2w^3+3w^2+4w-52=0
a little synthetic division yields
(w-2)(w^3 + 4w^2 + 11w + 26)
and no easy roots thereafter
(one irrational, two complex)
To find the roots of the given equation, we can use the factoring method or the quadratic formula. Let's start with the factoring method.
Step 1: Rearrange the equation in descending order of powers of w:
w^4 + 4w + 2^3 + 3^2 - 52 = 0
Step 2: Combine the constants:
w^4 + 4w + 8 + 9 - 52 = 0
w^4 + 4w - 35 = 0
Step 3: Try to factorize the quadratic expression. In this case, it is not easily factorizable, so we will proceed with the quadratic formula.
Step 4: Apply the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / (2a)
For our equation w^4 + 4w - 35 = 0, we have a = 1, b = 4, and c = -35. Substituting these values into the quadratic formula, we get:
w = (-(4) ± √((4)^2 - 4(1)(-35))) / (2(1))
w = (-4 ± √(16 + 140)) / 2
w = (-4 ± √(156)) / 2
w = (-4 ± √(4 * 39)) / 2
w = (-4 ± 2√(39)) / 2
w = -2 ± √(39)
Therefore, the roots of the equation w^4 + 4w - 35 = 0 are:
w = -2 + √(39) and w = -2 - √(39)