Question

this a little long question on quadratic inequalities. please go through my working and clarify my questions.

Answer 1

\[ \frac{ x^2+6x-11 }{ x+3 }>-1\]

Answer 2

the answer given is \[(-\infty,-8) U (-3,1)\]

Answer 3

\[\frac{ x^2+6x-11 }{ x+3 }-1>0 \\ \frac{ x^2+6x-11-1(x-3)}{x+3}>0 \\ \frac{ x^2+7x-8}{x+3}>0 \\ ( x^2+7x-8)(x+3)>0 \\ ( x-1)(x+8)(x+3)>0 \\ \textrm{we get the required answer}\]

Answer 4

\[\textrm {where as \in this question } \frac{ 3x^2-2x-5 }{ x^2-2x+5 }>2 \\ \textrm{the answer given is} (-5,3) \\ \textrm {which can be done like this} \\ 3x^2-2x-5 >2(x^2-2x+5) \\ or x^2+2x-15>0 \\or (x+5)(x-3)>0\]

Answer 5

so whats the doubt ?
(x+5)(x-3) >0 implies
x+5 > 0 , x-3 > 0 or x+5<0, x-3 <0
the first gives (-5,3)

x+5<0, x-3 <0 ---> x<-5, x>3 which is not possible (no solution for this)

Answer 6

\[\textrm {my question is why can't we solve \it like the way the previous \sum was solved } \\ \frac{ 3x^2-2x-5 }{ x^2-2x+5 }>2 \\\frac{ 3x^2-2x-5 }{ x^2-2x+5 }-2>0\\\frac{ 3x^2-2x-5 -2(x^2-2x+5) }{x^2-2x+5 }>0 \\(x^2+4x-15)(x^2-2x+5)>0 \\ \textrm {like this}\]

Answer 7

there are two problems which i have done, please go through both and give your reason. ..thanks in advance!!

Answer 8

\(\\\frac{ 3x^2-2x-5 -2(x^2-2x+5) }{x^2-2x+5 }>0 \\ \implies 3x^2 -2x-5-2x^2+4x-10>0\\ x^2+2x-15>0\)
exactly the same result you did the other way!

Answer 9

you'll get same results form both the ways

Answer 10

yes, thats incorrect, how u got that ?

Answer 11

\[\textrm {what about this part? is this incorrect ?}\\ \frac{ 3x^2-2x-5 -2(x^2-2x+5) }{x^2-2x+5 }>0 \\(x^2+2x-15)(x^2-2x+5)>0\]

Answer 12

the numerator is just x^2+2x-15

Answer 13

isn't this the way the first problem was solved ?

Answer 14

but in the first problem we did bring the denominator up by multiplying both sides with (x+3)

Answer 15

if we did not do that the answer to first problem would have been ( x-1)(x+8)>0

Answer 16

3x^2−2x−5−2x^2+4x−10>0
the 2 methods you showed are equivalent and will lead to same inequality in the end,
x^2+2x-15>0

Answer 17

what happens to the denominator ? in the second case ?

Answer 18

ok, wait, i can think of a reason why you *cannot* do like
\(3x^2-2x-5>2(x^2-2x+5)\)
because you can't multiply both sides by x^2-2x+5 as you don't know whether x^2-2x+5 >0 or < 0 so you wouldn't know whether to switch signs or not

Answer 19

the best thing to do is your second method

Answer 20

subtract 2 and bring it in the form of > 0

Answer 21

but in the second method what happens to the denominator ? \frac{ 3x^2-2x-5 -2(x^2-2x+5) }{x^2-2x+5 }>0

Answer 22

the numerator comes to x^2+2x-15 that is fine, so should we only factorize it to solve the inequality ?

Answer 23

then isn't this solution incorrect (the one below) ?

Answer 24

\[\frac{ x^2+6x-11 }{ x+3 }-1>0 \\ \frac{ x^2+6x-11-1(x-3)}{x+3}>0 \\ \frac{ x^2+7x-8}{x+3}>0 \\ ( x^2+7x-8)(x+3)>0 \\ ( x-1)(x+8)(x+3)>0 \\ \textrm{we get the required answer}\]

Answer 25

ok, we use this,
if a/b > 0
a>0, b>0 or a<0, b<0

Answer 26

ok

Answer 27

so the way i did it is not proper have to treat (x-1)(x+8)>0 and (x+3)> 0 separately !!

Answer 28

hmm, yeah

Answer 29

thanks :)