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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
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
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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