What is number of real solutions of x which satisfies (x-1)(2x+1)(x+1)(2x-3) = 15?

1 Answer
Sep 20, 2017

Only real roots of (x-1)(2x+1)(x+1)(2x-3)=15 are {2,-3/2}.

Explanation:

Let us group the LHS in (x-1)(2x+1)(x+1)(2x-3)=15 as follows

{(x-1)(2x+1)}xx{(x+1)(2x-3)}=15

or (2x^2-2x+x-1)(2x^2+2x-3x-3)=15

or (2x^2-x-1)(2x^2-x-3)=15

Now let us assume 2x^2-x=t, then above equation becomes

(t-1)(t-3)=15 or t^2-4t+3=15

or t^2-4t-12=0 or (t-6)(t+2)=0

Hence t=6 or t=-2

If t=6, then 2x^2-x=6 or 2x^2-x-6=0

Here discriminant is (-1)^2-4*2*(-6)=49 which is square of a ratioonal number and we can get rational roots. The above equation becomes

(2x+3)(x-2)=0 i.e. x=2 or x=-3/2

If t=-2, then 2x^2-x=-2 or 2x^2-x+2=0

Here as discriminant is (-1)^2-4*2*2=-15 and as it is negative, we do not have real roots.

Hence, only real roots of (x-1)(2x+1)(x+1)(2x-3)=15 are {2,-3/2}.