Quadratic game!

Put $\frac{ax^{2}-7x+5}{5x^{2}-7x+a}=y$ then $(a-5y)x^{2}-7x(1-y)+(5-ay)=0$ In order that the values of x found from this quadratic may be real, the expression $49(1-y)^{2}-4(a-5y)(5-ay)$ must be positive, that is, $(49-20a)y^{2}+2(2a^{2}+1)y+(49-20a)$ must be positive; Hence $(2a^{2}+1)^{2}-(49-20a)^{2}$ must be negative or zero, and (49-20a) must be positive. Now $(2a^{2}+1)-(49-20a)^{2}$ is negative or zero, according as $2(a^{2}-10a+25)\times2(a^{2}+10a-24)$ is negative or zero; That is, according as $4(a-5)^{2}(a+12)(a-2)$ is negative or zero.

This expression is negative as long as a lies between 2 and -12 , and for such values (49-20a) is positive; the expression is zero when a=5, -12 or 2 , but (49-20a) is negative when a=5 . Hence, the limitting values are 2 and -12 ,
and a may have any intermediate value.