#ALGEBRA



How many integral values can the expression (15x^2+2x+1)/(x^2-2x-1) not take?

Solution-

As we have to find the values that the equation cannot take,we can find range of possible values that equation will attain and then remove this range from R(R->Real numbers)

=>(15x^2+2x+1)/(x^2-2x-1) = y
=>(15x^2+2x+1)=y*(x^2-2x-1)
=>(15x^2+2x+1)=y*x^2-2yx-y
=>(15-y)x^2 +2*(y+1)*x + (1+y) = 0
Now for x to be real discriminant must be >=0
So,
=>(2^2)*(y+1)^2 - 4(15-y)(y+1) >=0
=>(y+1)[y+1-15+y]  >=0
=>(y+1)(y-7) >=0

So, y = (-inf,-1] u [7,inf)

So the given equation will not take integral values between (-1,7)
i,e; 0,1,2,3,4,5,6

Hence,the given equation cannot take 7 integral values. 

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