Latest Elitmus Aptitude Question SOLUTION: Number of possible sollutions of x for the equation (2x+9)^2=x+3 ?

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Number of possible sollutions of x for the equation (2x+9)^2=x+3 ?

Asked In Elitmus sujith (8 years ago)

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ajit kumar sharma  (8 years ago)

(2x+9)^2=x+3
=>4x^2+36x+81=x+3
=>4x^2+35x+78=0
now discriminant b^2-4ac=35^2-4*4*78=1225-1248=-23
here b^2-4ac
here Discriminant is negative...
ans will be no real solution ony two imaginary solutions......

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ajit kumar sharma  (8 years ago)

(2x+9)^2=x+3
=>4x^2+36x+81=x+3
=>4x^2+35x+78=0
now discriminant b^2-4ac=35^2-4*4*78=1225-1248=-23
here b^2-4ac

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Bikash kumar  (8 years ago)

i think question is find the number of value of x where (2x+9)^2/(x+3) value is integer
option a)2 b)4 c)6 4)infinity

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Kethineni Vinodh  (8 years ago)

x=0,-2,-4,6,-6,-12 total


x=6 possible solutions

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Mukesh  (8 years ago)

0 Because If You Simplify This Equation You Will Get 4x^2 +35x+78 = 0. And By The Rule b^2-4ac = -23. So, Answer Is 0 Real Solutions.

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prateek bhati  (8 years ago)

Since the equation is quadratic so the possible solution are 2 ,despite of either roots are real or imaginary.

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nitin dhami  (8 years ago)

2 real roots as the calculated delta comes out to be +ve

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Amit Sankhyan  (8 years ago)

Write (2x + 9 )^2 as (2x + 6 + 3) ^2 => ( 2(x +3) + 3 )^2 .
Now Square it: 4 ( x+ 3) ^2 + 9 + 12(x+3) .
Now divide this equation by x+3. ( as (2x+9)^2 = x+3 => (2x+9) / (x+3) =0 ) .
So divide each term by x+3.
Then take 9 / (x+3) [Do not consider other two terms as they can have infinite values for x] . So for this term to be real, x can have -12, and 6 as the value.
So 2 values of X are possible.

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