Question on quadratic polynomials with real cofficients

Question

Let$ P(x) = x^2 + ax + b$ be a quadratic polynomial with real coefficients. Suppose there are real number $s ≠ t$ such that $P(s) = t$ and $P(t) = s$. Prove that $ b - st$ is a root of the equation $ x^2 + ax + b - st = 0$.

Answer 1

You have:

$s^2 + as + b = t$ [eq. $1$]

$t^2 + at + b = s$ [eq. $2$]

You need to show that $(b-st)^2 + a(b-st) + b-st = 0$. The $LHS$ of the latter can be factorised to $(b-st)(b-st + a + 1)$, so it suffices to show that at least one of these factors is zero.

Now take equation $1$ minus equation $2$, factorise and rearrange:

$(s+t)(s-t) + a(s-t) + (s-t) = 0$

Cancel $(s-t)$ as $s \neq t$:

$a + 1 = -(s+t)$ [eq. $3$]

Assume (for now) that neither $s$ nor $t$ is zero (this case will be handled separately). Take equation $1$ times $t$ minus equation $2$ times $s$, factorise, rearrange:

$st(s-t) + b(t-s) = (t+s)(t-s)$

Again, cancelling $(s-t)$ gives:

$b = (s+t) + st$ [eq. $4$]

Substituting the expressions in equations $3$ and $4$ into $(b-st + a + 1)$, we get:

$b - st + a + 1 = (s+t) + st - st - (s+t) = 0$, and hence the proposition is proven.

The remaining case is when exactly one of $s$ or $t$ is zero (both cannot be zero since they are unequal). Because of the symmetry between $s$ and $t$, it suffices to consider just one case. Let $s=0$. Equation $1$ yields $b=t$. Putting that into equation $2$ gives $t^2 + at + t = 0$.

Now $b-st = t$, and the expression $(b-st)^2 + a(b-st) + b-st$ reduces to $t^2 + at + t$, which we've already shown to be zero. Hence the proposition holds true in the case where one of $s$ or $t$ is zero, as well.

Answer 2

Consider the polynomial,$$P_1(x)=x^2+(a-1)x+b$$ its roots are $s$ and $t$ then by theory of roots $b=st$ or $b-st=0$ and you are asking that $x^2+ax+b-st$ passes through origin.