Quadratics Solutions problem

Question

For what values of $m$ does $x^2−4x−m=0$ have no real solutions while $x^2−9x+m^2=0$ has at least one real solution?

Answer 1

Hint: Have a look at the discriminant $D=b^2-4ac$ for $f(x)=ax^2+bx+c$

If $D<0$ the equation $f(x)=0$ has no real solution. If $D\geq 0$ the equation $f(x)=0$ has at least one real solution.

Answer 2

take the first equation $$f(x)=x^2-4x-m$$ $$f'(x)=2x-4=0\rightarrow x=2$$ for real solution the $f(2)\geqslant 0$

and for no real solution if $f(2)<0$

For the second equation

for real solution the $f(4.5)\geqslant 0$

and for no real solution if $f(4.5)<0$

Answer 3

The first equation has no real solution provided $$(-4)^2-4(-m)=4(4+m)\lt 0,$$ while the second has at least one real solution provided $$(-9)^2-4m^2=3^2-(2m)^2\ge 0.$$

Both conditions are satisfied when both inequalities hold simultaneously. Can you continue now?