Math, asked by AryanBaranwal, 1 year ago

Hello Friend IIT Based question.
Solve for x:-
$|{3x}^{2} + 4x| = 1$
Answer it fast.

Answers

Answered by SaiSoniya

Here, |3x^2+4x|=1

==3x^2+4x=1 or 3x^2+4x=-1

==3x^2+4x-1=0 or 3x^2+4x+1=0

==x=(-4+-√(16+12))/6 or 3x^2+3x+x+1=0

==x=(-4+-√28)/6 or 3x(x+1)+1(x+1)=0

==x=2(-2+-√7)/6 or (3x+1)(x+1)=0

==x=(-2+-√7)/3 or 3x+1=0,x+1=0

3x=-1 , x=-1

x=-1/3 , x=-1

SaiSoniya: ok

SaiSoniya: as the question has modulus when we remove it we get one plus and one minus on the opposite side i.e on the right side

SaiSoniya: then we solved using factorisation

SaiSoniya: if it is not applicable then we used to solve for a value x

SaiSoniya: hi

SaiSoniya: sorry

SaiSoniya: I am not getting the equation to send u

AryanBaranwal: ok

AryanBaranwal: no problem

SaiSoniya: it's -b+-√(b^2-4ac)/2a

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