some1 pls help me with this question - x^2+3|x|+2=0,find the number of solutions?

Answer 1

Well the above solution is wrong.
The solution doesnt satisfy the equation.
Now the given equation can be written as |x|^2 + 3|x| + 2 =0
Hence |x| = -2 or -1
However |x| is always positive .
Hence the equation has no solutions.

Answer 2

if the equation is x^2+3|x|+2=0, there are no solutions because you must sole for the absolute value and you get |x|=(-x^2 - 2)/3. Any number squared is positive, so if you take the opposite of that, you get a negative. When you subtract 2 and divide by 3, the number is still negative and the absolute value can never equal a negative.
If the equation is -x^2+3|x|+2=0, then there are four solutions.
|x|=(x^2 - 2)/3
Because of the absolute value, write as 2 equations.
x = (x^2-2)/3 and -x = (x^2-2)/3
Multiply by 3 to get rid of the fraction.
3x = x^2 - 2 and -3x = x^2 -2
Write in standard form so you can factor.
x^2 - 3x - 2 =0 and x^2 +3x -2 = 0
Use the quadratic formula and get
[3+sqrt(17)]/2, [3-sqrt(17)]/2, [-3+sqrt(17)]/2, and [-3-sqrt(17)]/2,

Answer 3

I would sort into 2 parts

one part x >= 0 that is x^2+3x+2 = 0 => (x+2)(x+1) >0 x = -2 or -1
this is not meeting condition so no positive solution
now for x < 0
x^2-3x+2 = 0 or (x-2)(x-1) = o or x = 2 or 1 which is again not meeting condition
so there is no -ve solution
So there is no solution

Addiitonally this can be done in amother way
each term is non -ve so LHS >= 2
which can never be 0
so no solution

Answer 4

Supose x real. Then x^2>0 and |x|>0, hence x^2+3|x|+2>2, therefore x^2+3|x|+2 don't have real solutions.

Supose x not real. Then x^2 = -3|x|+2 is real (because |x| is real), hence x = ki (since x is not real), for some real k. Then

x^2+3|x|+2 = 0 implies -kk+3|k|+2 = 0.
If k>0 this implies -kk+3k+2=0, ie, k = (-3 +17^(1/2))/2
if k<0 then -kk-3k+2 = 0, ie, k = (3 -17^(1/2))/2

Then the equation have two solutions,
x = (-3 +17^(1/2))i/2
x = (3 -17^(1/2))i/2

Answer 5

In the given problem, you need to find all values of x which satisfy the equation,not |x|, hence the negative values of x cannot be ignored.

x^2+3|x|+2=0
Here,x can be +ve or -ve, because |-x|=x and (-x)^2=x^2
So, the resultant equation will remain
x^2+3x+2=0,irrespective of the sign of x.

x^2+3x+2=0
or,(x+-2)(x+-1)=0
or,x=+-2,=-1
Hence the equation will have 4 possible solutions.

Answer 6

x^2+3x+2=0

x^2+2x+x+2=0

x(x+2)+1(x+2)=0
(x+1)(x+2)=0

as product of two nos is 0
therefore either x+1=0 or x+2=0

hence the two possible answers are- x= -1 or x= -2

Answer 7

You need to make it bracketed again..

eg (x +- ) (x +- )

I havent filled it in because I don't know if that says -x squared, or just x squared.

In any case, the main idea is that if either bracket adds to zero, then the overall equation will be zero, because multiplication by zero yields zero.

That is important to know.

If you don't yet understand that, please add details now, and I will explain further...

Answer 8

first x^2+3x+2=0
=(x+2)(x+1)=0
so x=-2 or x=-1

second x^2-3x+2=0
=(x-2)(x-1)=
so x=2 or x=1

so thesolutions are
-2,-1 or 2,1

Answer 9

Karthik I is correct, there are no solutions. However, you would need to show there are no imaginary solutions as well, something his 'proof' doesn't demonstrate.

Answer 10

1 solution, no negative solutions because of the absolute vaule.