Solve the following for x
l (x+h) ^2 - x^2 l or equal to 3h^2, h0

Question

Solve the following for x
l (x+h) ^2 - x^2 l > or equal to 3h^2, h>0

anonymous · Answer

@wio  could you help me with this problem?

anonymous · Answer

$$
|(x+h)^2-x^2|\geq 3h^2
$$?

anonymous · Answer

yes

anonymous · Answer

First simplify inside: $$
(x+h)^2-x^2 = x^2+2hx+h^2-x^2 = 2hx+h^2
$$

anonymous · Answer

Next, we know $h>0\implies 3h^2>0$ so both are positive, meaning if we square them then the inequality will remain true.

anonymous · Answer

$$
(2hx+h^2)^2 \geq 9h^4
$$

anonymous · Answer

now do I just simplify that and solve for x?

anonymous · Answer

Yeah, you wanna solve for the $=0$ case.  You can use quadratic formula.

anonymous · Answer

When you know the points where it crosses the $0$ mark, you know when the equality changes between true and false.

anonymous · Answer

You'll end up with something along the lines of $$
(x-r_1(h)) (x-r_2(h)) \geq 0
$$Where $r_1$ and $r_2$ are roots, and they happen to be functions of $h$.

anonymous · Answer

oh okay got it, so do I do $$\left| x-1 \right|+\left| x-2 \right|<3 hint: consider \cases for x \le 1, 1<x \le 2, x>2$$

anonymous · Answer

???  No idea where you got that.

anonymous · Answer

sorry I meant to say, would I do that problem the same way I did the other one?

anonymous · Answer

that was a miswording from my side

anonymous · Answer

No, the other problem is completely different.

anonymous · Answer

ohh could you please explain how it's different? Just to know for future purposes

anonymous · Answer

It's not a quadratic equation.

anonymous · Answer

ohh how would approach this problem then?

anonymous · Answer

but basically, you have to consider when $|x-2|>0$ and $|x-3|>0$

anonymous · Answer

that is why they said to consider $x>2$ and $x>3$ and also $x\lt 2$

anonymous · Answer

whoops... should be $1$ and $2$... not $2$ and $3$.