Finding number of real roots

Question

The question is

Find the number of real roots of the equation $(x-1)x(x+1)(x+2)-1=0$

I simply tried to simplify the equation but it transforms into biquadratic which becomes really tough to conclude anything from it.My second approach was :- $y=(x^2-1) ;$ $y=\frac{1}{(x^2+3x+2)}$ Then i tried to find their intersection but i failed Please help to solve the question without using any tools (like wolfram) other than pure mathematics since it is subjective mathematical paper's question which needs a proper solution.

Answer 1

The function is symmetric about $x = -{1 \over 2}$ since the zeroes of $(x+2)(x+1)x(x-1)$ are at $-2,-1,0,$ and $1$. So letting $y = x + {1 \over 2}$ the polynomial is $$(y-{1 \over 2})(y + {1 \over 2})(y - {3 \over 2})(y + {3 \over 2}) - 1$$ $$ = y^4 - {5 \over 2}y^2 - {7 \over 16}$$ This is a quadratic equation in $y^2$, so you can find all possible $y$ easily, and from that the number of solutions $x$.

Answer 2

Idea:

Write $y=x(x+1)= x^2+x$ then you get $$(x-1)(x+2)= x^2+x-2 = y-2$$ ...

Answer 3

The equation $(x-1)x(x+1)(x+2)=0$ has four roots, at $x=-2,-1,0$ and $1$. Including the $-1$ pushes the root at $-2$ a bit to the left and the root at $1$ a bit to the right. What happens to the two roots in the middle?

They are pushed toward one another. However, it is possible that they are pushed exactly to eachother and become a single root. It is also possible that they are pushed past one another and vanish (become complex).

So we have to see what happens in this case. It is most easily done by inserting $x=-\frac12$ (the value exactly in between the two roots; the function is symmetric about this $x$ value, so it will let is conclude either way) and multiplying out. We get $$ \left(-\frac32\right)\left(-\frac12\right)\left(\frac12\right)\left(-\frac32\right)-1=\frac{9}{16}-1<0 $$ which is to say, the two roots were pushed past one another and vanished. So the equation has two roots.

Answer 4

Factorize

$(x^2+x)(x^2+x-2)-1=0$

$(x^2+x)^2-2(x^2+x)-1=0$

$(x^2+x-1)^2=2$