(x^3-6x-5)=0
2007-08-31 04:53:30 · 6 answers · asked by Jan1ce 1 in Science & Mathematics ➔ Mathematics
x^3-6x-5=0
by inspection,x=-1 is a root , so (x+1) is a factor of LHS. so
x^2(x+1)-x(x+1)-5(x+1=0
(x+1(x^2-x-5)=0
x+1=0,x=-1ANS
OR x^2-x-5=0
x=[1+/-sqrt(1+20)]/2
=[1+/-sqrt(21)]/2. ANS
2007-08-31 05:05:29 · answer #1 · answered by Anonymous · 0⤊ 0⤋
If you ment x^2 and a +5 then:
(x-5)(x-1)=0
The way you have it factors to:
x^2(x-6)-5=0 x^2(x-6) = 5 OR
(x+1)(x^2-x-5) = 0
x= -1 OR x=(1+- sqrt 21)/2
2007-08-31 12:05:31 · answer #2 · answered by 037 G 6 · 0⤊ 0⤋
(x+1)(x^2-x-5)=0
either x=-1 or x^2-x-5=0
2007-08-31 12:05:52 · answer #3 · answered by Anonymous · 0⤊ 0⤋
(x+1)(x^2-x-5)
2007-08-31 12:05:24 · answer #4 · answered by Snow 1 · 0⤊ 0⤋
x^3 - 6x - 5 = 0
x (x^2 - 6) - 5 = 0
Is it really x^3 or x^2?
2007-08-31 11:59:24 · answer #5 · answered by petep73 3 · 0⤊ 0⤋
are you sure it is x^3 and not x^2?
If it was x^2, you could use the quadratic formula.
2007-08-31 12:01:11 · answer #6 · answered by soelo 5 · 0⤊ 0⤋