Question

The product of two numbers is 1, and the difference of their squares is 3.75. Find the two numbers.

Answer 1

\[xy=1\iff y=\frac{1}{x}\]

Answer 2

solve
\[x^2-\frac{1}{x^2}=3.75\]

Answer 3

The system is :
\[xy=1\\x^2-y^2=3.75\]
From the 1st equation we can write :
\[y=\frac1x\]
So, the 2nd becomes :
\[x^2-\frac1{x^2}=\frac{15}4\]
Then :
\[4x^4-4-15x^2=0\]
Now let
\[t=x^2\]
the last equation becomes :
\[4t^2-15t-4=0\]
We have :
\[\Delta= 15^2-4\times 1\times(-4)=241\]
So we have :
\[t_1=\frac{15+\sqrt{241}}{8}\\
t_2=\frac{15-\sqrt{241}}{8}<0
\]
So we have :
\[x=\pm\sqrt{t_1}=\pm\sqrt{\frac{15+\sqrt{241}}{8}}\]
And the values of y are very clear !

Answer 4

thanks to the both of you :3