Question

Find the set of values of x for the following inequality

(x ^{2}+4x-3)div(x^{2}+1) 15 years ago

Answer 1

\[(x ^{2}+4x-3)\div(x^{2}+1)<x\]

Answer 2

http://www.wolframalpha.com/input/?i=%28x2%2B4x%E2%88%923%29%C3%B7%28x2%2B1%29%3Cx
this ill help you i believe:)

Answer 3

ok... is good... thx, but i need the steps...

Answer 4

\[\frac{x^2+4x-3}{x^2+1}-x<0\]
\[\frac{x^2+4x-3}{x^2+1}-\frac{x(x^2+1)}{x^2+1}<0\]
\[\frac{x^2+4x-3-x^3-x}{x^2+1}<0\]
x^2+1>0 so we can multiply both sides by x^2+1 and the inequality still holds
\[-x^3+x^2+3x-3<0\]
\[x^3-x^2-3x+3>0\] since we multiply by -1 on both sides the inequality flipped
now hopefully we can find some rational zeros to x^3-x^2-3x+3=y
lets try factor by grouping
\[x^2(x-1)-3(x-1)=(x-1)(x^2-3)=0=> x=1, x=\sqrt{3},x=-\sqrt{3}\]
it works!
now
we have to test the intervals (-inf,-sqrt{3}) and (-sqrt{3},1) and (1,sqrt{3}) and (sqrt{3},inf)
to see where y>0

Answer 5

you just choose a number in each of those intervals and where you have y>0
then that interval is apart of your solution

Answer 6

hats off to myininaya!
hope i could give you more medals!

Answer 7

i'm going to bed if neoh does not have any questions

Answer 8

at first i was gonna try synthetic division but i seen that factor by grouping might work and i was happy lol

Answer 9

---- +++ --- +++
-----|-----|-----|----
-sqrt{3} sqrt{3} 1

so {x:-sqrt{3}<x<sqrt{3} or x>1} is the ans??

Answer 10

wait 1 comes first then sqrt{3}

Answer 11

f(-2)=-3
f(0)=3
f(1.5)=-.375
f(4)=39

so (-sqrt{3},1) U (sqrt{3},inf)

Answer 12

--- +++ --- +++
-----|-----|-----|----
-sqrt{3} 1 sqrt{3}

so {x:-sqrt{3}<x<1 or x>sqrt{3}} is the ans??

Answer 13

yep!

Answer 14

ok... thx for the sloving the problem.. ^^

Answer 15

np

Answer 16

good night