A $p$ real parameter value determines how many roots the equation

$$
2 x^{2}-10 p x+7 p-1=0
$$

has in the interval $(-1 ; 1)$.
Solution. The discriminant of the polynomial

$$
D=100 p^{2}-8(7 p-1)=4\left(25 p^{2}-14 p+2\right)=4\left[\left(5 p-\frac{14}{10}\right)^{2}+\frac{4}{100}\right]>0
$$

indicates that the equation has solutions for all values of the parameter $p$.

Let's examine the function $f(x)=2 x^{2}-10 p x+7 p-1$ on the interval $(-1 ; 1)$. If $x=-1$, the function value is $f(-1)=1+17 p$; if $x=1$, then $f(1)=1-3 p$. The function changes sign if $f(1) \cdot f(-1)<0$, which is equivalent to $(1+17 p)(1-3 p)<0$. This inequality holds if $p<-\frac{1}{17}$ or $p>\frac{1}{3}$. 

If $-\frac{1}{17}<p<\frac{1}{3}$, then $f(1) \cdot f(-1)>0$, and both $f(1)=1-3 p>0$ and $f(-1)=1+17 p>0$. Notice that regardless of the value of $p$, $f(0.7)=-0.02$. This is negative, but $f(1), f(-1)>0$, so the function changes sign twice in this interval. Therefore, the equation has two roots in the interval $(-1 ; 1)$ if $-\frac{1}{17}<p<\frac{1}{3}$.

If $p=-\frac{1}{17}$, then $f(-1)=0$; if $p=\frac{1}{3}$, then $f(1)=0$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_6df3ba6650ec2c70d671g-1.jpg?height=553&width=585&top_left_y=916&top_left_x=759)

In summary: if $p \geq \frac{1}{3}$ or $p \leq-\frac{1}{17}$, the function has one root in the interval $(-1 ; 1)$, and if $-\frac{1}{17}<p<\frac{1}{3}$, it has two roots.