how many solutions does x^2 + 4lxl + 3 =0 has?

since every term in the given equation is positive, therefore neglecting modulus, we have-

x^2 +4x+3=0

solving by splitting middle term or by using quadratic formulae, we get-

(x+3)(x+1)=0

or, x=-3 and x=-1

putting these values in original equation we get-

24 and 8, which are no solutions to the given equation.

therefore given equation has no real solution.

enjoy,

AJ

we can solve the question by this method also, which i perfer is better. lets go this way-

we have- x^2+4|x|+3=0,

in this all the term are positive, and x^2 can be written as |x|^2, because both represent positive values.

therefore equation becomes-

|x|^2+4|x|+3=0

let |x| be t=a variable t,we have-

t^2+4t+3=0

then by solving by splitting middle term-

t^2+3t +t+3=0

or t=-3 and t=-1

i.e |x|=-1 and -3 , which is not possible, therefore no possible solutions.

Hope this helps

AJ