Question

find the solution of :

Answer 1

\[\sqrt{x^{2} - 3x - 10} - 2\sqrt{x^{2} - 3x - 9} + \sqrt{x^{2} - 3x - 8} < 0\]

Answer 2

i take u = x^2 - 3x - 10, give
sqrt(u) - 2 sqrt(u + 1) + sqrt(u + 2) < 0
sqrt(u) + sqrt(u + 2) < 2sqrt(u+1)
2u + 2 + 2sqrt(u^2 + 2u) < 4u + 4
2 sqrt(u^2 + 2u) < 2u + 2
4(u^2 + 2u) < 4u^2 + 8u + 16
o < 16
i was missing of u :D

Answer 3

Looks good !

Answer 4

i saw an answer through wolfram, but how ? there is no explanation

Answer 5

the solution is x<= -2 or x>= 5 (wolfram's version)
but i need a way

Answer 6

the 4th line can only happen if u>0 and u+2>0
the intersection of those sets are u>0
So we want u>0
which means x^2-3x-10>0 <--is what you need to solve

Answer 7

solving this will make your last line true
you know that 0<16

Answer 8

actually, i know just square both sides any times to get u and the rule that the domain in sqrt must be >= 0
sqrt(u) + sqrt(u + 2) < 2sqrt(u+1)
how about the right side ?

Answer 9

x^2-3x-10>=0

Answer 10

I don't understand the question you are asking @tanjung

Answer 11

and yes \[\sqrt{A} \cdot \sqrt{B}=\sqrt{A \cdot B } \text{ \iff } A \ge 0 \text{ and } B \ge 0\]

Answer 12

you did this in your fourth line

Answer 13

2 sqrt(u^2 + 2u) < 2u + 2
is that ?

Answer 14

\(\sqrt{u} - 2 \sqrt{u + 1} + \sqrt{u + 2} < 0\)
yields
\(0 < 16\)
there is no mistake in your work

but you also need to look at the domain of \(u\) :
\[u\ge 0 ,~~~u+1\ge 0,~~~u+2\ge 0\]

Answer 15

right

Answer 16

intersection of all those sets mentioned by rational is u>=0

Answer 17

so, we dont need find u then solve for x ?

Answer 18

your inequality 0<16 is true for u in the domain of the original equation

Answer 19

so 0<16 happens when u>=0
replace u with what you set it equal to
and solve that inequality for x

Answer 20

we do need to find \(u\) then solve for \(x\). find \(u\) from below set of inequalities :
\[0\lt 16,~~u\ge 0 ,~~~u+1\ge 0,~~~u+2\ge 0\]

Notice the first inequality is an useless inequality

Answer 21

I should have included u+1>=0 earlier
and also look at the original equation
I for some reason just looked at the sqrt(AB)=sqrt(A)*sqrt(B) part
:p

Answer 22

but anyways the intersection of all those sets in terms of u is still the same

Answer 23

oh, ok i think i can understand now. thanks for you both @myininaya and @rational :D