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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
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