19. The function $f(x)$ is defined as $f(x)=\frac{x-1}{x+1}$.
The equation $f\left(x^{2}\right) \times f(x)=0.72$ has two solutions $a$ and $b$, where $a>b$. What is the value of $19 a+7 b$ ?
SOLUTION
134

We are required to solve $\frac{x^{2}-1}{x^{2}+1} \times \frac{x-1}{x+1}=\frac{18}{25}$, which simplifies to $\frac{(x-1)(x-1)}{x^{2}+1}=\frac{18}{25}$.
This leads to the quadratic equation $7 x^{2}-50 x+7=0$ which factorises as $(7 x-1)(x-7)=0$. Therefore, $a=7$ and $b=\frac{1}{7}$ and $19 a+7 b=19 \times 7+7 \times \frac{1}{7}=133+1=134$.