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?

Recall the discriminant of quadratic equation: https://en.wikipedia.org/wiki/Discriminant — EBP, Jun 14 '19 at 18:49

callculus42 · Answer 1 · 2019-06-14T19:14:13.250

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.

edited Jun 14 '19 at 19:14

answered Jun 14 '19 at 18:53

callculus42

30,550

E.H.E · Answer 2 · 2019-06-14T19:19:48.680

0

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 $f(4.5)<0$

edited Jun 14 '19 at 19:19

answered Jun 14 '19 at 19:11

E.H.E

23,280

not clear OP knows calculus – gt6989b Jun 14 '19 at 19:14
@gt6989b However, this is another way to solve the problem – E.H.E Jun 14 '19 at 19:16

score 0 · Answer 3 · answered Jun 14 '19 at 20:01

0

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?

answered Jun 14 '19 at 20:01

Allawonder

13,327

I would strongly prefer for the OP to do some work on the problem instead of us solving everything for him. – gt6989b Jun 14 '19 at 20:04
@gt6989b I don't think your definition of everything agrees with the most accepted one, which I had assumed. – Allawonder Jun 14 '19 at 20:12
indeed, which is why i removed the -1 after some thought, but left the comment... let's hope the op ends up doing something after your hint... – gt6989b Jun 14 '19 at 20:59

Quadratics Solutions problem

3 Answers3