Parameter

[Loki123]

 

[topsquark]

View attachment 32606

Let me ask you this. What are the solutions if a = 1/2, for example. How many (legal) solutions do we get?

-Dan

 

[Loki123]

Let me ask you this. What are the solutions if a = 1/2, for example. How many (legal) solutions do we get?

-Dan

2

 

[Danteth]

2

What are they specifically?

 

[BigBeachBanana]

The statement [imath]a\ge 1[/imath] is false. Test out a few values.

 

[Loki123]

The statement [imath]a\ge 1[/imath] is false. Test out a few values.

okay, but in the answer i need it to go from [1,3/2) so how do i get it then??

 

[Loki123]

okay, but in the answer i need it to go from [1,3/2) so how do i get it then??

and here is wolfram alpha

Sqrt[2-x]=Divide[-x,2]+a - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

 

[Danteth]

When you write [math]-\frac{x}{2}+a \geqslant 0[/math] try to estimate x in terms of a and remember that [math]x \leqslant 2[/math] Then you get [math]-2a\leqslant 2[/math] So the statement [math]a \geqslant 1[/math] is correct, but it's only low estimation of a

 

[BigBeachBanana]

okay, but in the answer i need it to go from [1,3/2) so how do i get it then??

In the OP.
[imath]x \le 2[/imath] and [imath]x\le 2a \implies a\ge1[/imath]
Put your domains together.

 

[Loki123]

In the OP.
[imath]x \le 2[/imath] and [imath]x\le 2a \implies a\le1[/imath]
Put your domains together.

I am a bit confused how you combined those inequalitions.

 

[BigBeachBanana]

I am a bit confused how you combined those inequalitions.

If [imath]x=2[/imath] and [imath]x=2a[/imath], what is the value of a?

 

[Loki123]

If [imath]x=2[/imath] and [imath]x=2a[/imath], what is the value of a?

1
but
if x≤2
x could be -1

x≤2a
a could be 0 and it would still work

so is it safe to assume this:
2≤2a
cause it this case it would be wrong

 

[BigBeachBanana]

1
but
if x≤2
x could be -1

x≤2a
a could be 0 and it would still work

so is it safe to assume this:
2≤2a
cause it this case it would be wrong

I meant to say [imath]a \ge1[/imath], not [imath]a\le1[/imath]

 

[Loki123]

I meant to say [imath]a \ge1[/imath], not [imath]a\le1[/imath]

so if x is 0
x≤2

x≤2a
a could also be 0 zero

a≥1
so what now?

 

[BigBeachBanana]

so if x is 0
x≤2

x≤2a
a could also be 0 zero

a≥1
so what now?

If x=0, then 2*(something greater or equal to 1) is still greater than 0.

 

[Loki123]

If x=0, then 2*(something greater or equal to 1) is still greater than 0.

okay
so can i combine them like this
x≤2
x≤2a

2≤2a
1≤a

 

[topsquark]

okay
so can i combine them like this
x≤2
x≤2a

2≤2a
1≤a

Right. And a < 3/2, so [imath]1 \leq a < 3/2[/imath]. Done!

You missed the point of my earlier post. If a = 1/2 then the only possible solution is x = 2. That means there is only one solution. The problem requires two solutions. If a < 1 then you only get one solution.

-Dan