I don't know how to even start on this one, please help.
Find all real solutions.
(x^2 - x - 22) ^3/4 = 16
2007-09-04 02:29:12 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(x² - x - 22) ^3/4 = 16
x² - x - 16^4/3 - 22 = 0
x² - x - 62.3 = 0
(x - 8.4)(x + 7.4) = 0
x = -7.4, 8.4
2007-09-04 02:40:59 · answer #1 · answered by gebobs 6 · 0⤊ 0⤋
Just raise both sides to the power of 4/3, and then use the quadratic formula.
You should get:
x = 1/2 (1 +/- Sqrt[89 + 128 * 2^(1/3)])
where +/- means "plus or minus."
-G.
Edit: I wondered the same thing that RAJA posits: The problem would be much easier if it were "4/3" instead of "3/4." Go back and check to be sure.
.
2007-09-04 09:36:47 · answer #2 · answered by Anonymous · 0⤊ 0⤋
the question should be
(x^2 - x - 22) ^4/3 = 16
therefore,
(x^2 - x - 22) = 8
x^2 - x - 30 = 0
therefore, x = 6 and/or -5
2007-09-04 09:40:20 · answer #3 · answered by raja 2 · 0⤊ 0⤋
Raise both sides to power 4/3
(x^2-x-22)=16^(4/3)
16^(4/3)=40.32
x^2-x-22-40.32=0
x^2-x-62.32=0
This is of form ax^2+bx+c=0
a=1 b=-1 c=-62.32
x=[-b+-sqrt(b^2-4ac)]/2a two real roots
2007-09-04 09:44:37 · answer #4 · answered by cidyah 7 · 0⤊ 0⤋