Question

if |(x^2+2x+2)+(3x+7)| < |x^2+2x+2| + |3x+7|.
then set of complete values of x is

Answer 1

heck of a question !

Answer 2

You can use desmos.com/calculator and do it yourself ;)

Answer 3

I'll go the human way, :P , ok give me a moment.

Answer 4

Anyone can use a calculator @koalamon but that doesn't mean the person understands what they're doing. It also doesn't teach them the fundamentals for more advance mathematics courses. It's best if the person knows what they're doing. Good job @kappa007 for tackling the problem head on.

Answer 5

k, just a sec, someone's in urgent need. I'll be back in like 5 mins. I hope you can wait.

Answer 6

okk...

Answer 7

He didn't ask for anyone to explain, he asked to find the complete values of x, and the calculator requires you to put in the information in an understandable format. If it comes out as a graph you will need to isolate anywhere where y=0

Answer 8

okay, the quadratic expression here has a negative discriminant, do you agree ?

Answer 9

no ?

Answer 10

So, if the discriminant is negative, it means the quadratic expression doesn't cut the x-axis at all i.e, it is always positive or always negative.

Answer 11

ryt...

Answer 12

We can take 2 cases when \[x \ge \frac{ -3 }{ 7 }\]
and
\[x < \frac{ -3 }{ 7 }\]

Answer 13

Try it out. See what happens. Tell me if there's doubt.

Answer 14

I am not completely sure but how about using this |A+B| < |A| + |B| so either |A| <0 or |B| < 0 or both |A| and |B| <0 now if solve these inequality we reach our answer.

Answer 15

In this case, A remains always positive, so that's a help.

Answer 16

kappa007 how are u getting it as -3/7 i m getting it -7/3

Answer 17

yeah true we can write A as (x+1)^2+1...and @samigupta8 I also feel it should be -7/3

Answer 18

ohh , silly mistake, ignore it, you're right

Answer 19

that's obv -7/3

Answer 20

did you get the ans ?

Answer 21

okk...so now we will solve it like
for x<-7/3

|(x^2+2x+2)+(3x+7)| < x^2+2x+2-3x-7
(x^2+2x+2)+(3x+7) < x^2-x-5
x^2+5x+9<x^-x-5
6x<-14
x<-7/3

Answer 22

m i rytt...

Answer 23

so, that's you answer. When we assume the 2nd case, the result is in agreement. That itself is the required interval.

Answer 24

yup

Answer 25

and for case 1st it will all come out to be zero

Answer 26

one point, you can directly cancel out the quadratic exp on both sides before simplifying.

Answer 27

for case 1, it comes out 0 < 0 which is absurd, so we infer that the interval doesn't have any solution.

Answer 28

okk..thankuu...