24for max solutions...
x2+4x+3=√x+3
Squaring..
x4+8x3+6x2+23x+6=0
put x=-x
=> x4-8x3+6x2-23x+6=0
Using descartes rule of sign change
max zero positive roots
max 4 negative roots
20 Answers
24
62Yes this si fine..
by the original question, real roots should be greater than -3
Are you getting that..
Really sorry I din make too much sense of your post #9 ..
11What I did was this....
Since x=-3 is the obvious soln, we look for other solutions, thus we get the eqn
\sqrt{x+3}(x+1)=1
From which we have
(x+3)=\frac{1}{(x+1)^2}
Now put (x+3)=t.....BTW is the other root an extraneous one?
62how did you get that equation..
if you should have had a √t ...
to eliminate that you would probably have had a 4th degree equation!
11I'm in total confusion.....
From where i lrft in my last post, we can clearly see f(23)>0 and f(2)<0 - so that gives us 1 root over there.....Again f(3)>0, so that's 1 root in that interval from 2 to 3....So all in all 3 roots!!!!!!!
24hey arshad..
in ur post #7
it should be 2 intersection points in 2nd quadrant....
i know it was a typo form u
62I think graph kills it...
But why Bhargav gave it is to see if you can see
f(x) = f-1(x)
I think that is why this dicussion is moving for so long..
24i wrote that because i didnt find any relevance of this discussion stretching so long.....
becoz everything is clear from graph ,,,
so whats the real problem ???????????????????
39eureka , i donot find any relevance of your post here.
anyone would have done it. so wats the use in finding the max possibility? it leads us no where here .
24i am not saying all these will really exist...but these are max possibllty
11Edit :- t=3 gives a positive value for f(t), so that gives 3 solutions.....nowhere after that the derivative evanishes, thus no more than 3......roots.
11Well this qsn has 2 solutions.....
Firstly -3 is an obvious soln....
Then u have f(t)=t^3-4t^2+4t-1, where t=x+3
f'(t)=(3t-2)(t-2)
And from 2nd derivative test f(t) has a max at t=\frac{2}{3} and min at t=2, so f(t) must cross the axis once in between, thus 2 solutions......
1i drew the graph of
x^2 + 4x + 3 and
√(x+3)
from it
i got 2 intersection points in the 3rd quadrant
39will my confirmation give u the anser? u finding yourself will be the right spirit
anyways for the answer , no one yet?
i want explanations arshad and organic...........
1one solution at
x=-3
but i think this question has more to it than meets the eye....