Mathematics - Quadratic Equation Question with Solution | TestHub

Given $x^2+px-q=0$ has roots $\alpha ,\beta$.

So, $\alpha +\beta =-p$ and $\alpha \beta =-q$.

Given $x^2+px+r=0$ has roots $\gamma ,\delta$.

So, $\gamma +\delta =-p$ and $\gamma \delta =r$.

The expression is $\frac{(\alpha -\gamma )(\alpha -\delta )}{(\beta -\gamma )(\beta -\delta )}$.

We know that for a quadratic $a{x}^2+bx+c=0$ with roots $r_1,r_2$, we can write $a(x-r_1)(x-r_2)=a{x}^2+bx+c$.

So, $x^2+px+r=(x-\gamma )(x-\delta )$.

Substitute $x=\alpha$ into $x^2+px+r=(x-\gamma )(x-\delta )$:

${\alpha }^2+p\alpha +r=(\alpha -\gamma )(\alpha -\delta )$.

Since $\alpha$ is a root of $x^2+px-q=0$, we have ${\alpha }^2+p\alpha -q=0$, which implies ${\alpha }^2+p\alpha =q$.

Substitute this into the previous equation:

$q+r=(\alpha -\gamma )(\alpha -\delta )$.

Similarly, substitute $x=\beta$ into $x^2+px+r=(x-\gamma )(x-\delta )$:

${\beta }^2+p\beta +r=(\beta -\gamma )(\beta -\delta )$.

Since $\beta$ is a root of $x^2+px-q=0$, we have ${\beta }^2+p\beta -q=0$, which implies ${\beta }^2+p\beta =q$.

Substitute this into the previous equation:

$q+r=(\beta -\gamma )(\beta -\delta )$.

Therefore, $\frac{(\alpha -\gamma )(\alpha -\delta )}{(\beta -\gamma )(\beta -\delta )}=\frac{q+r}{q+r}=1$, provided $q+r\ne 0$.