View Full Version : How do i solve this?
Gughanath
Oct3-04, 08:12 AM
x^2+7x>6+2x^3...I tried many times..i cant solve this :grumpy: ...please help
Dr-NiKoN
Oct3-04, 08:27 AM
x^2 + 7x > 6 + 2x^3 = -2x^3 + x^2 + 7x - 6 > 0
Try solving for:
-2x^3 + x^2 + 7x - 6 = 0
Gughanath
Oct3-04, 08:37 AM
x^2 + 7x > 6 + 2x^3 = -2x^3 + x^2 + 7x - 6 > 0
Try solving for:
-2x^3 + x^2 + 7x - 6 = 0
I cant solve that sorry...could you help any further?
I noticed the sum of the coefficients is 1 so (x-1) is a factor:
2x^3-x^2-7x+6 = (x-1)(x+2)(2x-3)
That should get you started!
aekanshchumber
Oct6-04, 02:03 AM
try it,
x^2 + 7x>6x + 2x^3
or, 2x^3 - x^2 -x<0
or, x(2x^2 - x - 1)<0
this implies, either x<0 or (2x^2 - x - 1) <0
as the expression is negative which is only possible if one of the term is negative.
Integral
Oct6-04, 02:32 AM
try it,
x^2 + 7x>6x + 2x^3
or, 2x^3 - x^2 -x<0
or, x(2x^2 - x - 1)<0
this implies, either x<0 or (2x^2 - x - 1) <0
as the expression is negative which is only possible if one of the term is negative.
You have read the problem incorrectly. It should read
x^2 + 7x > 6 + 2x^3
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